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siric

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Posts posted by siric

    Simple Query with results on same page

    in PHP Coding Help

    Posted

    Try this.  As I do not have your CSS file, I cannot get the styling exact, but you should be able to get through that.  Also, define all of your classes in the CSS file and replace

     

    <p align="left" style="margin-top: -1; margin-bottom: -1">

     

    with

     

    <p class="whatever">

     

     

    //CSS file

.whatever {
text-align= left;
margin-top: -1;
margin-bottom: -1
}

     

     

    This will make your code easier to read and the formatting easier to change when ever you need to.

     

     

    <body>
	<?php
	 	if(!isset($_POST['submit'])){
	?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
	<p align="left" style="margin-top: -1; margin-bottom: -1">
	<span class="style2"><b><font size="2">Search by Zip:</font></b>
    <input type="text" name="zipCode" size="20"> 
	<b><font size="2">(example: 01922)  </font></b></span></p>
    <p align="left" style="margin-top: -1; margin-bottom: -1">
    <span style="font-family: Arial, Helvetica, sans-serif"><br>
    <font size="2"><b>Search by State:     </b></font> 
	<input type="text" name="state" size="20">
	<b> <font size="2">(example: NY for New York) </font></b></span>
	<input type="submit" name="submit" value="Go">
</form> 

	<?php 
		} else {
		// Page two, search and display 
$connect = mysql_connect("localhost", "root", "sysadmin");  
$db = mysql_select_db("test", $connect);  

$sql = "SELECT * FROM STOREDATA WHERE State = '".$_POST['state']."' OR ZIP_Code = '".$_POST['zipCode']."'";  

$result = mysql_query($sql);  
$rows = mysql_num_rows($result);  
if($rows > 0) {  
	  echo "<p><center></p>
			<table border=\"1\">  
			<th align=\"center\newstyle\">Name</th> 
			<th align=\"center\">City</th> 
			<th align=\"center\">State</th> 
			<th align=\"center\">Sales Phone</th>"; 

	while($row=mysql_fetch_array($result)) {  
		echo "<tr><center>
				<td align=\"center\" class=\"newstyle\">".$row['Name']."</td>
				<td align=\"center\" class=\"newstyle\">".$row['City']."</td>
				<td align=\"center\" class=\"newstyle\">".$row['State']."</td>
				<td align=\"center\" class=\"newstyle\">".$row['Sales_Phone']."</td>
			  </tr>";  
	}  
}  
else {  
	echo "No results found";  
}  
mysql_close($connect); ;
		}
	?>
</body>

    Simple Query with results on same page

    in PHP Coding Help

    Posted

    Very possible.

     

     

     

    if (isset(submit)) {  //This checks to see if the submit button is set

//process query

} else {

//Show form  .......
<FORM name="search"  action="<?php echo $_SERVER['PHP_SELF']; ?> "  method="post" >  //This calls itself when the submit if clicked.

...........

     

     

    }

    this form validation is not doing anything

    in PHP Coding Help

    Posted

    There is no need to js.  In the <form> line use

     

    <FORM name="feature" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >

     

    This will cause the page to reload itself.

     

    Also where do you load the variables user->credits which is used in this line ?

     

      if($user->credits < '1000') {

    Strange posting error

    in PHP Coding Help

    Posted

    Tested this and it works.

     

     

    
<script type='text/javascript'>
  function disp_confirm(companyid) {
    var c = confirm('Do you really want to archive this company?');
if (c == true)  {
    window.location.href  = 'companies.php?action=Archive&companyid='+companyid;
}
  }
</script>




<?
$companyid = '1';
echo "<input type='image'  src='blue.png'   onclick=\"disp_confirm('$companyid')\" />";
?>

     

     

    Strange posting error

    in PHP Coding Help

    Posted

    Ok,

     

    Have done some further research and have come up with this so far.  The main problem seems to be the image for the form button.  If you do not use an image and use window.location, it works.

     

    function disp_confirm(companyid) {
var c = confirm('Do you really want to archive this company?');
      
      if (c == true)  {
            var link='companies.php?action=Archive&companyid='+companyid;
            window.location=link;

          //could also be
          // window.location = 'companies.php?action=Archive&companyid='+companyid;

      }

}

     

    and html

     

    echo "<input type='button' value='Redirect' onclick=\"disp_confirm('$companyid')\">";

     

     

    Will chk further on the image as button option.

    Strange posting error

    in PHP Coding Help

    Posted

    I moved around the code a bit as

     

    function disp_confirm(companyid) {
      var c = confirm('Do you really want to archive this company?');
      
      if (c == true)  {
            var link='companies.php?action=Archive&companyid='+companyid;
            alert(link);
      }
}

     

    The alert shows that the correct url is passed to link variable.  It would be just a matter of getting that into the URL.

    Problems with populating a combobox

    in PHP Coding Help

    Posted

    You need to find out how many rows the query has returned and then loop through the results.

     

     

    Pseudo Code

    
$number_of_rows = mysql_num_rows($sql_eventos)  //Get number of rows returned 

<select>

for ($count=0; $count< $number_of_rows; $count++) {
       <option value........>
}

</select>

    simple INSERT INTO problem

    in PHP Coding Help

    Posted

    Break down

     

    $query = sprintf ("INSERT INTO album (date, name, desc) VALUES ('$date', '$name', '$desc')");
    $result = mysql_query ($query);  //  Her stopper det tipper jeg

     

    to

     

    $query = "INSERT INTO album (date, name, desc) VALUES ('$date', '$name', '$desc')";

print $query;

$result = mysql_query ($query);  //  Her stopper det tipper jeg

     

    This will allow you to look at the query statement to ensure that it is correct and to run it directly against the database.  Also, why are you using sprintf ?

    Calling each id from mysql

    in PHP Coding Help

    Posted

    Ok,

     

    Went through it in detail and this works.  Don't even need arrays - my bad.

     

    
$max_number_of_cats = 10;

for ($count = 1; $count < $max_number_of_cats+1; $count++) {  //start number for count is the first cat number

    $sql = "SELECT * FROM photo WHERE photo_category='$count' ORDER BY photo_id DESC";
    $result= mysql_query($sql);
$num_rows = mysql_num_rows($result);
    	
	if ($num_rows > 0) {   //If there is a result for the cat search, then display the divs
	//Start div here

	for ($i=0; $i<$num_rows;$i++) {
		$photo_id=mysql_result($result, $i, "photo_id");
		$photo_caption=mysql_result($result, $i, "photo_caption");
		$photo_filename=mysql_result($result, $i, "photo_filename");
		$photo_category=mysql_result($result, $i, "photo_category");
		$photo_colour=mysql_result($result, $i, "photo_colour");

		print "Photo - $photo_id - $photo_caption - $photo_filename - $photo_color - $photo_category<br/>";
         
	}
print "Cat Change<br/><br/>";  //end your div here
}
}

     

    Try that and let me me know.

    Calling each id from mysql

    in PHP Coding Help

    Posted

    You would need to use an array for the categories.

     

    Find out how many categories you have and define an array -

     

    $cat = array();

for ($count = 1; $count < $max_number_of_cats+1; $count++) {

      $result = mysql_query( "SELECT photo_id,photo_caption,photo_filename FROM art_photos WHERE photo_category='".addslashes($cat[$count])."' ORDER BY photo_id DESC" );
      //cat[$count] here would equate to cat1 in the first instance, cat2, etc.



      if ($result) {   //If there is a result for the cat search, then display the divs
          ...
           //make div and display
         ...
     }
}

     

     

    Insert Date from PHP form into MySQL date column

    in PHP Coding Help

    Posted

    Whenever I get into problems with variables, I just validate them by assigning $_POST variables to a normal var and printing them.

     

    So as Muddy_Funster has suggested, assign the POST to $date and do a print afterwards to ensure that it is correct.

     

    Also, note that your entry field allows free text, so unless you are using this yourself, it can lead to problems, as users can enter the date in any format.

     

     

    change file extension

    in PHP Coding Help

    Posted

    I am probably WAY off mark here, but wouldnt this also work?

      $file_name = $file_name_array['0'];
  $new_name = str_replace($substr($file_name, -3, 3), 'flv', $file_name);

     

    You do not even have to place the name in an array to do it this way.

     

    Simply

    $file_name = 'myvideo.avi';
$new_name = str_replace(substr($file_name, -3, 3), 'flv', $file_name);

     

    That would work.  The limitation here is that it will only work with file extensions of 3 characters in length.  And remember, no $ in front of the substr function name.

    Submit button dont work

    in PHP Coding Help

    Posted

    Works for me.  What I would do however is to use the same name for the submit button, but only change the value

     

    <input type="submit" name="submit" id="tilbake" value="Back" />
<input type="submit" name="submit" id="lagre" value="Finished" />
<input type="submit" name="submit" id="publiser" value="Publish" />

     

    Then in the preview.php

     

    $selection = $_POST['submit'];

if ($selection == 'Back') {

...

} elseif ($selection == 'Finished') {

.....

} elseif ($selection == 'Publish') {

......

}

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