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fj1200

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Posts posted by fj1200

    DB Update page appears blank but it was working... I have no idea why.

    in PHP Coding Help

    Posted

    I think I aught to clarify a bit - when the user clicks a button on the main overview page to end the job, this page is opened with a text field to insert the final count.  They then enter said final count and submit, which THEN updates the database. 

     

    The problem I'm having is that when the user clicks the button on the main page, this page opens up but is blank.  So it hasn't even attempted the db connection at that time - it's something in the code itself and I can't see it.  Might just be a rogue ';' or " or something else but I can't find it.

    DB Update page appears blank but it was working... I have no idea why.

    in PHP Coding Help

    Posted

    mssql_connect() is ok and not the issue since (as I've said) I'm connecting to SQL Server 2000 and not MySQL in this instance  (http://uk2.php.net/manual/en/function.mssql-connect.php).

     

    The database connection was working fine - all I've done is to add the validation to an existing and working page.

     

    I'll see if I can get the errors but not convinced that's the problem.

     

    [edit]

    Nope - no errors generated.

    [/edit]

    DB Update page appears blank but it was working... I have no idea why.

    in PHP Coding Help

    Posted

    :-\ I put this in the MS SQL forum without realising.  Probably better here...

     

    Anyone got any ideas why this just gives me a blank page?

     

    Admittedly I took some of it from an example in a book, but the example worked and when I transfered the function to this is fails but I need to verify that they enter numbers ONLY - knowing the people involved on the client end they'll try it on...

     

    It may not b the function at all.  I've checked all me syntax, semicolons, braketing, etc.  All looks fine.  All it needs to do is enter a typed in final count, a timestamp and change a status flag in a SQL Server database.  It did work fine,  but I don't know what stopped it working.

     

    <!DOCTYPE HTML PUBLIC "-//W3C//Ddiv id=\"td\"  HTML 4.0 Transitional//EN">
<html xmlns="http://www.w3.org/1999/xhtml">
   <meta content="MSHTML 6.00.2900.2180" name="GENERATOR">
   

    <head><link rel="stylesheet" href="http://mrserver/css/mr.css" type="text/css" media="screen">
   <META HTTP-EQUIV="Refresh" Content="30;URL=http://mrserver/mr/mr_status.php">
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Status v1 - End Job</title>
    </head>
   <body>
      <?php

                    //Validate count to numbers only and max of 6 digits
                        function validateCount($count)
                        {
                            $regexp='/^([0-9]){1,6}$/';

                            if (eregi($regexp,$count)) return 1;
                                else return 0;
                        }

                     //Check form has been submitted...

                        if (isset($_POST['submit']))
                            {
                            $count=htmlentities($_POST['count']);
                            
                            if(validateCount($count))
                               // printf("%s ok", $count);

                        //Split POSTed job data into component parts
                            $job_end = explode("+", $_REQUEST[change_status]);

         //UPDATES Start data after theyhave ended the job
            $con2 = mssql_connect('dbserver', 'UID', 'PWD');
            if (!$con2 || !mssql_select_db('DB', $con2))
            {
                die('**** Unable to connect or select database! ****');
            }
               mssql_query("UPDATE JOB_MR 
                                             SET STATUS = '2', MR_END = GETDATE(), MR_COUNT = '$count'
                                             WHERE ID='$job_end[5]'",$con2)or die;
            mssql_close($con2);
                             else
                                printf("%s invalid", $count);
                            }
      ?>
<div style="font: 16px Verdana;" id="header">End Running Job</div><br>
        <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
            <input type="text" id="count" name="count" maxlength="6">
            <input type="submit" name="submit" id="submit" value="Submit Final Count">
        </form>
   </body>
</html>

     

    As a seperate question - what's the best IDE for PHP?  For the last 2 years I've used Notepad++ and I really like it, and am very comfortable with it but I've started using NetBeans and Eclipse - really so I don't get dependant on one system and also to see what else is available.  Sadly - it needs to be a freebe as the company won't buy anything like this.

    DB Update page was working and now appears blank but no idea why

    in Microsoft SQL - MSSQL

    Posted

    Anyone got any ideas why this just gives me a blank page?

     

    Admittedly I took some of it from an example in a book, but the example worked and when I transfered the function to this is fails but I need to verify that they enter numbers ONLY - knowing the people involved on the client end they'll try it on...

     

    It may not b the function at all.  I've checked all me syntax, semicolons, braketing, etc.  All looks fine.  All it needs to do is enter a typed in final count, a timestamp and change a status flag in a SQL Server database.  It did work fine,  but I don't know what stopped it working.

     

    <!DOCTYPE HTML PUBLIC "-//W3C//Ddiv id=\"td\"  HTML 4.0 Transitional//EN">
<html xmlns="http://www.w3.org/1999/xhtml">
<meta content="MSHTML 6.00.2900.2180" name="GENERATOR">


    <head><link rel="stylesheet" href="http://mrserver/css/mr.css" type="text/css" media="screen">
<META HTTP-EQUIV="Refresh" Content="30;URL=http://mrserver/mr/mr_status.php">
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Status v1 - End Job</title>
    </head>
<body>
	<?php

                    //Validate count to numbers only and max of 6 digits
                        function validateCount($count)
                        {
                            $regexp='/^([0-9]){1,6}$/';

                            if (eregi($regexp,$count)) return 1;
                                else return 0;
                        }

                     //Check form has been submitted...

                        if (isset($_POST['submit']))
                            {
                            $count=htmlentities($_POST['count']);
                            
                            if(validateCount($count))
                               // printf("%s ok", $count);

                        //Split POSTed job data into component parts
                            $job_end = explode("+", $_REQUEST[change_status]);

		//UPDATES Start data after theyhave ended the job
			$con2 = mssql_connect('dbserver', 'UID', 'PWD');
			if (!$con2 || !mssql_select_db('DB', $con2))
			{
			    die('**** Unable to connect or select database! ****');
			}
		   	mssql_query("UPDATE JOB_MR 
                                             SET STATUS = '2', MR_END = GETDATE(), MR_COUNT = '$count'
                                             WHERE ID='$job_end[5]'",$con2)or die;
			mssql_close($con2);
                             else
                                printf("%s invalid", $count);
                            }
	?>
<div style="font: 16px Verdana;" id="header">End Running Job</div><br>
        <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
            <input type="text" id="count" name="count" maxlength="6">
            <input type="submit" name="submit" id="submit" value="Submit Final Count">
        </form>
</body>
</html>

     

    As a seperate question - what's the best IDE for PHP?  For the last 2 years I've used Notepad++ and I really like it, and am very comfortable with it but I've started using NetBeans and Eclipse - really so I don't get dependant on one system and also to see what else is available.  Sadly - it needs to be a freebe as the company won't buy anything like this.

    $_POST not releasing variable value in web page

    in PHP Coding Help

    Posted

    Seems to be working now - not entirely 100% why - but there's a strange thing happening too: in the button code I now have:

     

    VALUE=\"$_job\" onclick=\"window.open('http://manu001/m_status/mr_end.php?job=$_job

     

    Changed the VALUE to $_job and it's working - I'd missed that so thanks for seeing the obvious!.  Was concerned it wouldn't display right but it's ok.  However - if I remove $_job from either the Value= or the href job=$_job it displays nothing.  I have to have both.  No idea why, but it's working now anyhow and it doesn't cause a problem.  I think the value= was causing the issue but still don't know why only one particular job was being picked up each time.

    $_POST not releasing variable value in web page

    in PHP Coding Help

    Posted

    It's a bit more involved but basically, yes.  The table sits in a form field set to POST, and one of the table columns has a hidden field that carries $_job as the value.  If I echo that back on the popup, it's all there as I'd expect it to be, just not the right job!  And it's always the same job that get's displayed no matter what  button I use, like it's not releasing it or it's picking up the wrong information from somewhere.    It's the only thing now that's not working.

    $_POST not releasing variable value in web page

    in PHP Coding Help

    Posted

    I'm having a bit of a problem with a web form.  It shows production status in one of our manufacturing areas with jobs listed in a table pulled from our production database - the job name appears on a button with a coloured background showing the job status - green for running and red for stopped.  The idea is that when the job is completed they hit the button on a touch screen an a confirmation window pops up and they fill in other data before confirming, which then updates a SQL Server db and closes the popup. 

     

    It all works as it should BUT the job name is always the same, like it's not releasing it.  I've tried setting it to "" at the end of the insert in the pop-up, or on the parent page before the variables are set, but no good.  It's the only part that isn't working and it's very frustrating.

     

    Not sure if I've got the button link to the popup right either.  It's generated on the fly and sits in the forst cullumn of a table. in a table

     

    echo "<input style=\"font: 16px Arial; width:140px; height:40px;\" name=\"change_status\" type=\"submit\" VALUE=\"$job_desc\" onclick=\"window.open('http://manu001/m_status/mr_end.php?job=$_job,'popup','width=550,height=550, etc....

     

    The variable $_job is a concatenated string of 10 other components that I then explode in the popup.  Just makes it a bit easier for me. 

     

    Anyone got any suggestions?  Probably something pretty simple.

    ajax noob - drop-down list populated fom live database when adding rows to page

    in Javascript Help

    Posted

    I have now got a working ajax drop-down, does everything I need EXCEPT - I can't get it to sit in the first cell.  Tried everything I can think of, tried loads of other people's ideas, just not working. 

     

    			function printpage(){
			window.print(); }
			//-->


		function insRow()
		{

		var ajaxRequest;  // The variable that makes Ajax possible!
		var mytable=document.getElementById("advicenote")
		for (var i=0; i<1; i++){ //add 3 new rows with content to the end of a table
		var newrow=advicenote.insertRow(-1) //add new row to end of table

		ajaxRequest.open("GET","ajax-advicenote.php", true);
		ajaxRequest.send(null); 

		ajaxRequest.onreadystatechange = function(){
		if(ajaxRequest.readyState == 4){

		var ajaxDisplay = document.getElementById('job');
		ajaxDisplay.innerHTML = ajaxRequest.responseText;
	}
}
ajaxRequest.open("GET","ajax-example.php", true);
ajaxRequest.send(null); 
}
			}
		var newcell=newrow.insertCell(0) //insert new cell to row
		newcell.innerHTML= ajaxRequest.responseText;

     

     

    I'm at a bit of a loss as to know what to do.  Can I give the newcell an id so I can reference it and so get the ajax response into it?  My first time with  ajax so be nice.  Taken me all day pretty much to get this far!

    ajax noob - drop-down list populated fom live database when adding rows to page

    in Javascript Help

    Posted

    I have this posted on the Javascript page but wonder if ajax is the solution.  I've been playing with it this morning but getting nowhere.

     

    function insRow()
         {
         var mytable=document.getElementById("advicenote")
         for (var i=0; i<1; i++){ //add 3 new rows with content to the end of a table
         var newrow=advicenote.insertRow(-1) //add new row to end of table
         
         var newcell=newrow.insertCell(0) //insert new cell to row
         newcell.innerHTML='<SELECT name="JOBNAME"><?php
               while($nt=mssql_fetch_array($result3))
               {
               echo "<option value='".$SITE_ID.".".$nt['JOB_NAME'].".".$nt['JOB_DESCRIPTION'].".".$JOBTYPE."'>".$nt['JOB_DESCRIPTION']."</option>";
               } ?></select>';
         
         var newcell=newrow.insertCell(1) //insert new cell to row
         newcell.innerHTML='<input name=\"depot_edition">';
         var newcell=newrow.insertCell(2) //insert new cell to row
         newcell.innerHTML='<input name=\"copies_bundle\">';
         var newcell=newrow.insertCell(3) //insert new cell to row
         newcell.innerHTML='<input name=\"bundles_std\">';
         var newcell=newrow.insertCell(4) //insert new cell to row
         newcell.innerHTML='<input name=\"bundles_odds\">';
         var newcell=newrow.insertCell(5) //insert new cell to row
         newcell.innerHTML='<input name=\"pallet_count\">';
         }
         }
      </script>

     

    The above code adds a row of input boxes to a page.  BUT - the first box is a drop-down list of running jobs that needs to be populated from the SQL Server production database at time of clicking the Add button.  Tried all sorts, but it's not working.  The query works fine though.  I'm a bit of a noob to JS and Ajax, so willing to learn.  Been reading Tutorials all morning. 

     

    Any help gratefuly recieved,

    Add a row and select dropdown with JS into php page

    in Javascript Help

    Posted

    Done some more digging: This almost gives me what I need:

     

    			function insRow()
		{
		var mytable=document.getElementById("advicenote")
		for (var i=0; i<1; i++){ //add 3 new rows with content to the end of a table
		var newrow=advicenote.insertRow(-1) //add new row to end of table

		var newcell=newrow.insertCell(0) //insert new cell to row
		newcell.innerHTML='<SELECT name="JOBNAME"><?php
				while($nt=mssql_fetch_array($result3))
				{
				echo "<option value='".$SITE_ID.".".$nt['JOB_NAME'].".".$nt['JOB_DESCRIPTION'].".".$JOBTYPE.\"'>".$nt['JOB_DESCRIPTION']."</option>";
				} ?></select>';

		var newcell=newrow.insertCell(1) //insert new cell to row
		newcell.innerHTML='<input name=\"depot_edition">';
		var newcell=newrow.insertCell(2) //insert new cell to row
		newcell.innerHTML='<input name=\"copies_bundle\">';
		var newcell=newrow.insertCell(3) //insert new cell to row
		newcell.innerHTML='<input name=\"bundles_std\">';
		var newcell=newrow.insertCell(4) //insert new cell to row
		newcell.innerHTML='<input name=\"bundles_odds\">';
		var newcell=newrow.insertCell(5) //insert new cell to row
		newcell.innerHTML='<input name=\"pallet_count\">';
		}
		}
	</script>

     

    But it's just the first <select> not being populated that's missing now.  Is it because it's not running the query at page load for the JS items since they are loaded later and on demand?  Any way round that? 

     

    I need to then enter all this into our database.  If I view the page source though, the new input fields don't show up.  Also, if the fields have the same name how do I cycle through them to build a db insert?  Not done this before, new territory so a little confused but getting there.

    Add a row and select dropdown with JS into php page

    in Javascript Help

    Posted

    I'm re-writing some of our PHP/MySQL-driven dispatching app cos the way we send out products is changing.  I now need to add multiple products onto one advice note whereas before they went out on one truck each and was easy.  I have a small js section on the advice note form that inserts rows, and that works.  Sort-of.  The problem is with the innerHTML part and adding the formatting of the cells which is controlled by css with <td id="input">.  How do I get the new cells to match the existing row formatting? 

     

    The table row has <input> tags in each cell except the first cell which has a drop-down <select>.  This is pulled of our live production database and just lists the current jobs.  it has a query:

     

    echo   "<td><select name=JOBNAME>";
		while($nt=mssql_fetch_array($result3))
		{
		echo "<option     value='".$SITE_ID.".".$nt['JOB_NAME'].".".$nt['JOB_DESCRIPTION'].".".$JOBTYPE."'>".$nt['JOB_DESCRIPTION']."</option>";
		}
echo	"</select>";

     

    How can I embed that into the JS?  I've tried putting it between <? ?> tags but it didn't work.  I'm not familiar with JS so this is all new to me.  I then need to add all the all the data in the new rows into a MySQL database and use that to print the advice note.  Still got  a lot of work to do on the back end, but I want to get this working first.

     

    It runs as linked page in Sharepoint and the guys that use it in our loading bay ore not the greatest with computers, so I need to make it as easy as possible for them.  And I have no other resources except me.  And I'm not a programmer or developer. 

     

    I've scoured the web, spent time trying it on my own with limited success, but now confused.

    Confused with choice of 2 Arrays not working...

    in PHP Coding Help

    Posted

    Sasa's one worked - in a test output I got

     

    FOLDER = 401 so Tower = uni --- foreach (uni as $key=>$tower)0 => T1 = 15 --> 0

    1 => T2 = 16 --> 0

    2 => T3 = 17 --> 0

    3 => T4 = 18 --> 0

    4 => T5 = 19 --> 0

    5 => T6 = 20 --> 0

    6 => T7 = 21 --> 0

    7 => T8 = 22 --> 0

    8 => T9 = 23 --> 0

    9 => T10 = 24 --> 0

     

    ..and so I can do stuff with that.    Ignace - sorry - your one produced the same error, but they've both helped me a lot.  Never come across variable variables before - but then I'm no web programmer, just a sysadmin, but learning and using php to write tools for the job - and enjoying it (mostly).

    Confused with choice of 2 Arrays not working...

    in PHP Coding Help

    Posted

    Can anyone assist with this?    I need the code to select between 2 arrays and then display the results accordingly but it's not working...

     

    	$arr_tower = "";

    if ($FOLDER == "101") $arr_tower = ('$kba');
if ($FOLDER == "201") $arr_tower = ('$kba');
if ($FOLDER == "301") $arr_tower = ('$uni');
if ($FOLDER == "401") $arr_tower = ('$uni');

echo 'FOLDER = '. $FOLDER . ' so Tower = '. $arr_tower;
echo ' --- foreach ('.$arr_tower.' as $key=>$tower)';

$kba = array(0=>"U1/2", 2=>"T2A", 3=>"U3", 4=>"U5/6", 5=>"T7");
$uni = array(0=>"T1", 1=>"T2", 2=>"T3", 3=>"T4", 4=>"T5", 5=>"T6", 6=>"T7", 7=>"T8", 8=>"T9", 9=>"T10"); 

echo "<table>";
foreach ($arr_tower as $key=>$tower)
{
$k = key($arr_tower); 
$val = $key + '15';
echo '<tr><td>'. $key .' => '. $tower. ' = '. $val .' --> '. $pbupdate[$val].'</td></tr>';
};
echo "</table>";

     

    Got to be something simple.  If I just use one array and the array name rather than $arr_tower it works fine.  (btw - I'm adding 15 to $key because the db query result for this section starts at $pbupdate[15] and it works.)

     

    This bit I added in as a test to see what exactly was being produced- 

    echo 'FOLDER = '. $FOLDER . ' so Tower = '. $arr_tower;
echo ' --- foreach ('.$arr_tower.' as $key=>$tower)';

    - produces the correct string...

     

    FOLDER = 401 so Tower = $uni --- foreach ($uni as $key=>$tower)

     

    ...however the web page shows this error:

     

    "Warning: Invalid argument supplied for foreach() in C:\wamp\www\ProdBoard\pb_job_update.php on line 245"

     

    ...which is the actual foreach() line.    I've tried moving the arrays to the top of the section; to where they are now; all oevr the place...  is there something I'm missing here?

     

     

    Set a form dropdown list 'selected' item to the value in a database

    in PHP Coding Help

    Posted

    OK - I've found a way to do it using arrays from another site (lost the link now) but it's a bit long-winded:

     

     

      $MACHINE= $pbupdate[4];
		$MODE = $pbupdate[5];

		$_machine = array(
		101=> "A",
		102=> "B",
		103=> "D",
		104=> "E",
		);

		If ($MACHINE == "101")  $PRESS = ("A");				
		If ($MACHINE == "201")  $PRESS = ("B");
		If ($MACHINE == "301")  $PRESS = ("D");
		If ($MACHINE == "401")  $PRESS = ("E");

	echo "<tr><td>Machine :</td><td>$PRESS</td><td>";

		echo "<SELECT name=fldr>";
		foreach ($_machine as $key => $value)
		{
		$SELECTED = "";		
		If ($value == $PRESS)  $SELECTED = ("SELECTED");
		echo '<OPTION '.$SELECTED.' value="'.$key.'">'.$value.'</option>';
		}
	echo "</select></td></tr>";

     

    It's a bit laborious for one field but it works, the only problem I have with it is I have 18 fields to update from drop-down boxes and 4 text fields.  Going to be a long file.

     

    If anyone has a simpler or more elegant solution I'd be glad to hear it.

    Set a form dropdown list 'selected' item to the value in a database

    in PHP Coding Help

    Posted

    Hi - I am building a form to update records in a database but struggling a bit with one secondary but vitally important area.  I wrote the data entry for using drop-down lists where the value is not the same as the list entry - for example, I have material widths where they are represented in the database as: 1/2 = 0, 3/4 = 1 and full width = 2.  There's 10 of these.  I have a machine number where A = 101, B=102, D=103, E=104 - this is how they are represented in the manufacturers DB and I'm following their naming for continuity and reporting. 

     

    I want to update these on the fly, so if, for example machine D is entered in the database, how do I set the "selected = ..." value in the drop-down on the form to reflect the database entry?

     

    eg.

    <Select>
<option  Value = \"101\">A</option>
<option Value = \"201\">B</option>
<option selected = \"selected\" Value = \"301\">D</option>  <--- like this...
<option  Value = \"401\">E</option>
</select>

     

    I only need to update maybe a few items - maybe we change the scheduled time, or put a job on a different machine, or the order changes - that kind of thing.  I don't want them to have to re-enter the whole job so there's a fair number of these to do on the update page.

     

    Been tearing my hair out and could do with some advice.

    [SOLVED] MySQL query displaying 1 row less in Php page

    in MySQL Help

    Posted

    Yup - that's got it.  I was using a &GET[] to populate the report title row but only using one element  of it then using a query to retrieve the rest - I now see the error of my ways (hangs head in shame) and use the $GET[] to provide the report title and header info (job id, name, date etc) and a MySQL query to get the detail.

     

    Many thanks - I was staring at it for a couple of hours yesterday.  What a dipstick!

    [SOLVED] MySQL query displaying 1 row less in Php page

    in MySQL Help

    Posted

    Odd thing happening here.  I have a couple of production reports that have got problems with a MySQL query.  I run the qry in Query browser - great, all data shows.  I run it on a report on a PHP page and I get the first row missing.  It started when I began to use <div>'s and css a few months ago but haven't had the time to revisit it, but I've been asked to do some more reports based on it - they like the look and feel of them - so I need to fix it.  Anyone else seen this?  I'm doing the queries EXACTLY as I would in a table, just substituting the table for div's.  I could rebuild it with a static page but I like how it looks and want to crack this issue.  Is it a css issue?  I have lots of static reports pages but really want to move to a more dynamic system.  I've started using PEAR as well although not here.

     

    It's not a browser thing - same in IE, FF, Chrom etc.

    CSS fails on php page when $_GET data used in URL

    in PHP Coding Help

    Posted

    Title says it all really.  CSS works fine when I use the page url on it's own - ok so I've got no data but I'm just making a working page look pretty and so it looks the same as all my other pages.  .

     

    However, it's called as a link from a list of jobs in a parent page pulled from a DB query, and when I run it from there - clicking the link - I get all the data but no css formatting.  Same results in any browser.  I already have a very similar page that works perfectly with the same css sheet and I can't see any differences.  All the quotes are there, all the ;'s and ()'s are there.  The only thing I can think of is that with the GET data it thinks it's looking for a different page?  Probably talking rubbish here though.

     

    Anyone able to help?

    [SOLVED] PEAR::HTML_AJAX component refresh

    in Other Libraries

    Posted

    81 views an nobody can help?

     

    Right - well, spent a few frustrating days digging around - can anyone tell me why this produced a page of garbage when it should be a .png image of a graph?  (btw - the timer bit works fine now)

     

    <html>
<head>
	<script type="text/javascript" src="../Includes/server.php?client=all"></script>

	<div id="target"></div>
		<script type="text/javascript">
	function graph()
	{
	HTML_AJAX.replace('target','../graphs/graph_stacked_rs.php');
	}

	function refreshpage()
	{
	setInterval(graph(), 10000);
	}
</script>
</head>
 <body  onload = "refreshpage();">

 </body>
   </html>

     

    It produces this:

     

    �PNG IHDR,�E��XIDATx���P���/�i�\*"���1�Wcp4s� G���Z��q��V{�r֩��8t���ǐZf��D�0ۨe)R�)���m�� ��$߄�_=99����_�r��큁����ݻw?y���ӓ������9_����,[���bŊ������gϞ������:H�Z�����J���o�]�h����u/& ��߿����y��尰0�B�q��˗�̦���۷O9v�X@@���K@@�'�v�����P��kמ;wΜ���'''�>}J�t钓�ӡC�<==�,Yr��=�}��{��]�z���ˢE��l���c��o߾��흞

     

    Should I be formatting it and saving it as a php page?  But if all it's importing is a .png from a PEAR IMAGE_GRAPH module (which works really well on normal php pages) I don't see why it should need to be in php format.  Or am I wrong there....

     

    Can anyone help please?

    [SOLVED] PEAR::HTML_AJAX component refresh

    in Other Libraries

    Posted

    I'm going to to be replacing the iframes on a php production dashboard with an ajax-driven page using the PEAR HTML_AJAX module but as yet there's no end-user documentation, and hunting the web - and this site - has not really revealed much.  I'm fairly new to PEAR, and have done very little JS but recently had my eyes opened to what AJAX could do for us.

     

    The iframes show graphs and datagrids driven by PEAR modules.  Just recently 'discovered'  AJAX,  but how do I get the individual components to refresh?  They point to different databases, some refresh once a minute, others I want to do once a day or once a week.  Sorry if it sounds a bit thick, but I've been looking for ages.

    [SOLVED] Run ncftpput from php and log the exit code

    in PHP Coding Help

    Posted

    Yup - that's got it.  why, oh WHY did I not see that??  ???

     

    Many thanks thorpe.

     

    This works:

     

    <?php

$return = -1;
$cmd = '"D:\Program Files\NcFTP\ncftpput" -u linktest -p linktest 192.168.2.2 / C:\wamp\www\Linktest\nft.txt';

exec($cmd,$output,$return);
if ($return == "0") 	$return = ("Success");
if ($return == "1") 	$return = ("Could not connect to remote host.");
if ($return == "2") 	$return = ("Could not connect to remote host - timed out.");
if ($return == "3") 	$return = ("Transfer failed.");
if ($return == "4") 	$return = ("Transfer failed - timed out.");
if ($return == "5") 	$return = ("Directory change failed.");
if ($return == "6") 	$return = ("Directory change failed - timed out.");
if ($return == "7") 	$return = ("Malformed URL.");
if ($return == "8") 	$return = ("Usage error.");
if ($return == "9") 	$return = ("Error in login configuration file.");
if ($return == "10") 	$return = ("Library initialization failed.");
if ($return == "11") 	$return = ("Session initialization failed.");

echo $return;
?>

     

    Now I can do some interesting stuff with it.

     

     

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