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witham

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    displaying pdf

    in PHP Coding Help

    Posted

    Problem solved!!

    After much more digging and reading I managed to do this by including 

    header('Content-Type: application/pdf');

     

    at the very top of the script and to display the file

     

    // Display uploaded.pdf
readfile($upfile);

     

    as the last function call.

    Thanks anyway I hope this helps someone else!!

    displaying pdf

    in PHP Coding Help

    Posted

    Thanks

    <html>
<head>
  <title>Uploading...</title>
</head>
<body>
<h1>Uploading file...</h1>
<?php

  if ($_FILES['userfile']['error'] > 0)
  {
    echo 'Problem: ';
    switch ($_FILES['userfile']['error'])
    {
      case 1:  echo 'File exceeded upload_max_filesize';  break;
      case 2:  echo 'File exceeded max_file_size';  break;
      case 3:  echo 'File only partially uploaded';  break;
      case 4:  echo 'No file uploaded';  break;
    }
    exit;
  }

  // Does the file have the right MIME type?
  if ($_FILES['userfile']['type'] != 'application/pdf')
  {
    echo 'Problem: file is not pdf text';
    exit;
  }

  // put the file where we'd like it
  $upfile = 'C:/uploads/'.$_FILES['userfile']['name'];

  if (is_uploaded_file($_FILES['userfile']['tmp_name'])) 
  {
     if (!move_uploaded_file($_FILES['userfile']['tmp_name'], $upfile))
     {
        echo 'Problem: Could not move file to destination directory';
        exit;
     }
  } 
  else 
  {
    echo 'Problem: Possible file upload attack. Filename: ';
    echo $_FILES['userfile']['name'];
    exit;
  }


  echo 'File uploaded successfully<br><br>'; 

  // reformat the file contents

  $contents = strip_tags($contents);
  $fp = PDF_open_file($upfile, 'w');
  fwrite($fp, $contents);
  
?>
</body>
</html>

    displaying pdf

    in PHP Coding Help

    Posted

    Hi I have written an upload script that restricts users from uploading anything other than pdf documents. After the php handles the upload and sends it to the directory I would really like to show a preview of the pdf that was uploaded. I have searched through the php.net website and have got really confused as to which pdf function to call. I assumed it would simply be PDF_open_file?? but I can't seem to get it to work.

    Before I continue it would be great to be certain that I am using the correct function.

     

    Thanks

    cannot connect to server

    in MySQL Help

    Posted

    Hi I hope someone can help I have been searching the net for hours looking for a solution. I have mysql running through xampp on a laptop running windows vista.

    I also have the identical setup on a pc running xp which works perfectly. My issue is that whilst I can connect to phpmyadmin through localhost I cannot connect ant databases that have been imported from the xp setup.

    The error I get is below, I have tried to echo the Host and password ect but they don't appear?

     

     

    Could not connect to database server:

\n"); // mysql_select_db selects a database to use on the database server //pointers to by $dblink // the @ sign before the command supresses any error messages @mysql_select_db ('weld' , $dblink) or die ("

Could not connect to the database
\n"); echo 'Host: '.$dbhost.'
'; echo 'Username: '.$dbuser.'
'; echo 'Password: '.$dbpass.'
'; echo 'Database: '.$dbname; ?>

Error Processing Query

     

    As the php works perfectly on xp setup I can't determine where the problem is??

    uploading file paths

    in PHP Coding Help

    Posted

    Thanks for you reply I have a database that contains various information input via a form this includes name, address and suchlike.

    Below is the array I want to display but I need the "compfile" to display as a hyperlink to the file.

    {


// $row["fieldname" returns the content for the field in the current row

   	echo "<tr><td>" .$row["MANNAME"].  "</td>";

   	echo "<td>"   .$row["prodtype"].  "</td>";

   	echo "<td>"   .$row["proddesc"].  "</td>";

echo "<td>"   .$row["prodnar"].	"</td>";

echo "<td>"   .$row["prodequiv"]."</td>";

echo "<td><a href=>"   .$row["compfile"]."</td></tr>";



}


// close html table tag
echo "</table>\n";

     

    uploading file paths

    in PHP Coding Help

    Posted

    hi i am hoping someone can help me, I need to upload the path to a pdf file stored on the same machine. I have tried "file upload" but I can't seem to make it work. I will continue to try but I just need to establish if this is the correct piece of code to be using?

    Cannot connect to server

    in MySQL Help

    Posted

    I hope someone can help I have set up xampp from apache friends on a laptop and written a php & mysql database which works fine.

    I loaded the same software, databases and php scripts onto another computer giving it the ip 192.168.0.5

    but it will not connect:

     

    These are the variables I am using to try and connect:

     

    $dbuser  = '';		// your database server user name
$dbhost  = '192.168.0.5';		// name of the sql server
$dbpass  = 'root';		// your database server password
$dbname  = 'weld';		// the name of the database to connect to

     

     

    and this is the mysql_connect and mysql_select_db

     

    // mysql_connect connects to the database server and returns a link to the the resource
$dblink = @mysql_connect("$dbhost","$dbpass") 
or die("<p><b>Could not connect to database server: ($dbhost)</b></p>\n");

// mysql_select_db selects a database to use on the database server pointers to by $dblink
// the @ sign before the command supresses any error messages
@mysql_select_db ($dbname , $dblink)
or die ("<p><b>Could not connect to database ($dbname)</b></p>\n");

     

    File navgation

    in PHP Coding Help

    Posted

    Hi I have a web application I am building and I would really like to navigate to a file so that this file path is subsequently rendered in a php script table as a link to the file.

    I have been reading alot about file upload but I don't think this is what I need as I only want to show the link where the file is and not actually upload it!

     

    This is the existing script:

     

    </select></td></tr>

<tr><td class="style8">Product Name:</td>
<td class="style8"><input type text name="ProdName"></td></tr>
<tr><td class="style8">Product File Path:</td>
[color=red]<td class="style8"><input type text name="FilePath"></td></tr>[/color]
<tr><td class="style8">Product Equivalent:</td>
<td><input type text name="ProdEquiv"></td></tr>

     

    The FilePath is the variable I want to pass to the script that handles the displaying of the link:

     

    {

// $row["fieldname"] returns the content for the field in the current row

   	echo "<tr><td>" .$row["MANNAME"].  "</td>";

   	echo "<td>"   .$row["prodtype"].  "</td>";

   	echo "<td>"   .$row["FilePath"].  "</td>";

echo "<td><a href=\"http://localhost/".$row["Compsheet"].".pdf\">".$row["Compsheet"]."</a></td>";

   	echo "<td>"   .$row["prodequiv"].  "</td></tr>";



}


// close html table tag
echo "</table>\n";

     

    I would be grateful if anyone could give me some assistance.

     

    Thanks

    Creating dynamic links

    in PHP Coding Help

    Posted

    Hi I am hoping someone out there can help me with the final piece to a php script I am creating, let me explain.

     

    I have a page that allows users to query a database using two drop down boxes populated dynamically the page that processes the results of this query shows the data in a table. One of the columns of this table contains data which I really want to be links to files

     

    This is what I have presently but the link doesn't work when I click it??

     

     

    // mysql_fetch_array fetches the results from the query a row at a time using a while loop each time it's called
// the result is returned as an array that can be referenced either by field name or by it's index
while ($row = mysql_fetch_array ($result)) 
// loop through the rows outputing them as html table rows


{

// $row["fieldname"] returns the content for the field in the current row

   	echo "<tr><td>" .$row["MANNAME"].  "</td>";

   	echo "<td>"   .$row["prodtype"].  "</td>";

   	echo "<td>"   .$row["proddesc"].  "</td>";

echo "<td><a href>"   .$row["Compsheet"].	"</td>";

   	echo "<td>"   .$row["prodequiv"].  "</td></tr>";



}


// close html table tag
echo "</table>\n";

     

     

    Thanks

    file path advice

    in PHP Coding Help

    Posted

    Thank you for your reply, I would really like the user to be able to view the pdf but everything I have read suggests that putting actual files into a database is not efficient as there could be alot? If the pdf could be embedded without compromising performance that would be fantastic

    file path advice

    in PHP Coding Help

    Posted

    I am sure this is possible but unsure exactly how to do it so let me explain.

     

    I would like to have a number of pdf files stored in a directory on the server.

     

    I would then like to include the paths to these files within a php array to output into an html table.

     

    Could anyone point me in the right direction please I have searched the forum and the net but nothing seems what I need??

     

    I thought perhaps include() might help but before I go ahead I thought I'd ask for help.

    Page rendering problem

    in PHP Coding Help

    Posted

    Thanks very much for the reply can I ask what you mean by the opening bracket, still can't see it (sorry)??

     

    I don't know if errors are turned on how would I check this please?

     

    Could you tell me why the opening and closing brackets may be an issue?

     

     

    Page rendering problem

    in PHP Coding Help

    Posted

    I would really appreciate a fresh pair of eyes looking at this code as it will not render properly it just displays the combobox at the bottom of the page. I have stared and stared had a break and stared and stared again but I cannot see the problem. I know it is something stupidly simply but "not seeing the wood for trees" springs to mind.

     

     

     

    <?

//check the $ProdName variable to see if it has data if not send the user back to the entry page
// this has to be done before any output

$ProdName = $_GET['ProdName'];
$ProdEquiv = $_GET['ProdEquiv'];
$Manufacturer = $_GET['Manufacturer'];
$ProdNar = $_GET['ProdNar'];
$Category = $_GET['Category'];


if (empty($ProdName)){  
header("Location: addprod.php"); 
die ("opps");  
}

if (empty($ProdEquiv)){  
header("Location: addprod.php"); 
die ("opps");  
}


$dbuser  = 'test';		// your database server user name
$dbhost  = 'localhost';		// name of the sql server
$dbpass  = '';		// your database server password
$dbname  = 'weld';		// the name of the database to connect to 

// use a compare statement to check for duplicates

$sqlCOMPARE	= "SELECT * from prodname where '$ProdName' = proddesc;";


// the sql query to insert data into the prodname table


$sqlINSERT	= "INSERT into prodname values
	  (null, '$Manufacturer', '$Category', '$ProdName', '$ProdNar', '$ProdEquiv');";

$sqlINSERT	= strtoupper($sqlINSERT);


//use to display the desired data $sqlSELECT

$sqlSELECT	= "SELECT MANNAME, prodtype, proddesc, prodnar, prodequiv 
	FROM man, produse, prodname where manno = manid
	and manid = '$Manufacturer'
	and useid = '$Category' 		   
	and prodid = useid
                		and manno = MANID
	and useid = USEID
	order by MANNAME;";

// mysql_connect connects to the database server and returns a link to the the resource
$dblink = @mysql_connect("$dbhost","$dbuser","$dbpass") 
or die("<p><b>Could not connect to database server: ($dbhost)</b></p>\n");

// mysql_select_db selects a database to use on the database server pointers to by $dblink
// the @ sign before the command supresses any error messages
@mysql_select_db ($dbname , $dblink)
or die ("<p><b>Could not connect to database ($dbname)</b></p>\n");


// now execute the next query to determine if a duplicate has been entered and use
// an if statement to return the user to the entry page if this is true


$result  = mysql_query($sqlCOMPARE, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");


if (  mysql_num_rows($result) > 0  ){  
header("Location: alreadyin.php");  
die ("opps");  

// now execute the next query to update the database table


$result  = mysql_query($sqlINSERT, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");

// now execute the next query to display these results

$result  = mysql_query($sqlSELECT, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");

?>


<head>
<title>W.E.L.D.</title>
</head>

<body>
<?

// output the table and the first row headings

echo'<table align = "center" border=0>' . "\n";
echo'<tr><td colspan = 3 align = "center"><table width="100" border="0" cellspacing="10" cellpadding="10">
  <tr>
    <th scope="row"><img src="car.jpg" width="150" height="150"></th>
    <td><img src="gears.jpg" width="150" height="150"></td>
    <td><img src="wheel.jpg" width="150" height="150"></td>
    <td><img src="metalworking.jpg" width="150" height="150"></td>
    </tr>
</table></td></tr>';



echo '<table align = "center" border="1" cellspacing="2" cellpadding="2">' . "\n";
echo "<tr bgcolor = #000000><td>MANUFACTURER</td><td>PRODUCT TYPE</td><td>PRODUCT NAME</td><td>PRODUCT DESCRIPTION</td><td>PRODUCT EQUIVALENT</td></tr>\n";


// mysql_fetch_array fetches the results from the query a row at a time each time it's called
// the result is returned as an array that can be referenced either by field name or by it's index


while ($row = mysql_fetch_array ($result)) 

// loop through the rows outputing them as html table rows


// $row["fieldname"] returns the content for the field in the current row


echo "<tr'><td>" 	. $row["MANNAME"]. "</td>";
   	echo "<td>" 	. $row["prodtype"]. 	"</td>";
   	echo "<td>" 	. $row["proddesc"]. 	"</td>";
echo "<td>"	. $row["prodnar"].	"</td>";
   	echo "<td>" 	. $row["prodequiv"]. 	"</td></tr>";}


// close html table tag
echo "</table>\n";

// the mysql_free_result command removes any resources relating to the query results
// this happens automatically at the end of the script but still better to free up now
mysql_free_result ($result);


// the mysql_close command severs the link to the database, with scripts that make multiple
// queries on the same database the command only needs to be done once after all queries are completed
@mysql_close ($dblink)     
or die( "<p><b>Error while closing connection to database server:" . 
        "($dbhost)</b></p>");

?>

<style type="text/css">
<!--

.combobox {
background-color: #000000;
color: #808080;
font-size: 12pt;
font-family: arial;
font-weight: bold;
}

-->
</style>

<form>
<table align=center border="0" width="80" cellspacing="0">
<tr><td width="100%" bgcolor="#000000">


<table border="0" width="100%" cellspacing="0" cellpadding="0">

</table>

<!-- put in a link returning the user to the original page -->
<table border="0" width="100%" cellspacing="0" cellpadding="3">
<tr><td width="2%">

<select class="combobox" name="SiteMap" onchange="if(options[selectedIndex].value){location = options[selectedIndex].value}" size="1">
<option value="">PLEASE CHOOSE AN OPTION</option>
<option value="querydatabase.php">SEARCH BY MANUFACTURER & PRODUCT TYPE</option>
<option value="addprod.php">ADD A PRODUCT</option>
<option value="addman.php">ADD A MANUFACTURER</option>
<option value="delprod.php">DELETE A PRODUCT</option>
<option value="delman.php">DELETE A MANUFACTURER</option>
<option value="searchweld.php">SEARCH FOR A PRODUCT BY NAME</option>
<option value="amendprod.php">AMEND A PRODUCT</option>
<option value="addcategory.php">ADD A CATEGORY</option>
<option value="selectequiv.php">AMEND AN EQUIVALENT</option>
</select>
</body>
</html>

    table row colour

    in PHP Coding Help

    Posted

    I am hoping someone can point me in the right direction, I have a php script that adds a row to an sql database but I would like to change the background colour of the added row:

     

    <?

//check the $ProdName variable to see if it has data if not send the user back to the entry page
// this has to be done before any output

$ProdName = $_GET['ProdName'];
$ProdEquiv = $_GET['ProdEquiv'];
$Manufacturer = $_GET['Manufacturer'];
$ProdNar = $_GET['ProdNar'];
$Category = $_GET['Category'];


if (empty($ProdName)){  
header("Location: addprod.php"); 
die ("opps");  
}

if (empty($ProdEquiv)){  
header("Location: addprod.php"); 
die ("opps");  
}


$dbuser  = 'user';		// your database server user name
$dbhost  = 'localhost';		// name of the sql server
$dbpass  = '';		// your database server password
$dbname  = 'ace;		// the name of the database to connect to 

// use a compare statement to check for duplicates

$sqlCOMPARE	= "SELECT * from prodname where '$ProdName' = proddesc;";


// the sql query to insert data into the prodname table


$sqlINSERT	= "INSERT into prodname values
	  (null, '$Manufacturer', '$Category', '$ProdName', '$ProdNar', '$ProdEquiv');";

$sqlINSERT	= strtoupper($sqlINSERT);


//use to display the desired data $sqlSELECT

$sqlSELECT	= "SELECT MANNAME, prodtype, proddesc, prodnar, prodequiv 
	FROM man, produse, prodname where manno = manid
	and manid = '$Manufacturer'
	and useid = '$Category' 		   
	and prodid = useid
                		and manno = MANID
	and useid = USEID
	order by MANNAME;";

// mysql_connect connects to the database server and returns a link to the the resource
$dblink = @mysql_connect("$dbhost","$dbuser","$dbpass") 
or die("<p><b>Could not connect to database server: ($dbhost)</b></p>\n");

// mysql_select_db selects a database to use on the database server pointers to by $dblink
// the @ sign before the command supresses any error messages
@mysql_select_db ($dbname , $dblink)
or die ("<p><b>Could not connect to database ($dbname)</b></p>\n");


// now execute the next query to determine if a duplicate has been entered and use
// an if statement to return the user to the entry page if this is true


$result  = mysql_query($sqlCOMPARE, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");


if (  mysql_num_rows($result) > 0  ){  
header("Location: alreadyin.php");  
die ("opps");  

}

// now execute the next query to update the database table


$result  = mysql_query($sqlINSERT, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");



// now execute the next query to display these results

$result  = mysql_query($sqlSELECT, $dblink)
or die("<p>Error Processing Query</p><hr /><p>".mysql_error()."</p>\n");

?>



<style type="text/css">
tr {color: white; font-family: arial; font-weight: bold;}
body {
background: navy;
background-image: url();
background-color: #333333;
}
.style2 {font-weight: bold; font-family: "arial", "Tempus Sans ITC", "Trebuchet MS";}
.style3 {font-family: Arial, Helvetica, sans-serif}
body,td,th {
color: #999999;
font-family: arial, Tempus Sans ITC, Trebuchet MS;
}
a:link {
color: #666666;
text-decoration: none;
}
a:visited {
text-decoration: none;
color: white;
}
a:hover {
text-decoration: underline;
}
a:active {
text-decoration: none;
}
a {
font-family: arial, Times New Roman, Times, serif;
color: #FFFFFF;
}
h1,h2,h3,h4,h5,h6 {
font-family: arial;
}
.style5 {font-family: arial, "Times New Roman", Times, serif}
.style7 {font-family: "arial", "Tempus Sans ITC", "Trebuchet MS"}
</style>


<head>
<title>W.E.L.D.</title>
</head>


<!-- set the css format for the page-->


<body>

<?

// output the table and the first row headings

echo'<table align = "center" border=0>' . "\n";
echo'<tr><td colspan = 3 align = "center"><table width="100" border="0" cellspacing="10" cellpadding="10">
  <tr>
    <th scope="row"><img src="car.jpg" width="150" height="150"></th>
    <td><img src="gears.jpg" width="150" height="150"></td>
    <td><img src="wheel.jpg" width="150" height="150"></td>
    <td><img src="metalworking.jpg" width="150" height="150"></td>
    </tr>
</table></td></tr>';



echo '<table align = "center" border="1" cellspacing="2" cellpadding="2">' . "\n";
echo "<tr bgcolor = #000000><td>MANUFACTURER</td><td>PRODUCT TYPE</td><td>PRODUCT NAME</td><td>PRODUCT DESCRIPTION</td><td>PRODUCT EQUIVALENT</td></tr>\n";


// mysql_fetch_array fetches the results from the query a row at a time each time it's called
// the result is returned as an array that can be referenced either by field name or by it's index


while ($row = mysql_fetch_array ($result)) 


// loop through the rows outputing them as html table rows

   {



// $row["fieldname"] returns the content for the field in the current row


echo "<tr><td>" . $row["MANNAME"]. 	"</td>";
   	echo "<td>" 	. $row["prodtype"]. 	"</td>";
   	echo "<td>" 	. $row["proddesc"]. 	"</td>";
echo "<td>"	. $row["prodnar"].	"</td>";
   	echo "<td>QUALUBE " 	. $row["prodequiv"]. 	"</td></tr>";


   }


// close html table tag
echo "</table>\n";

// the mysql_free_result command removes any resources relating to the query results
// this happens automatically at the end of the script but still better to free up now
mysql_free_result ($result);

// the mysql_close command severs the link to the database, with scripts that make multiple
// queries on the same database the command only needs to be done once after all queries are completed
@mysql_close ($dblink)     
or die( "<p><b>Error while closing connection to database server:" . 
        "($dbhost)</b></p>");

?>

     

    It is a while since my php college course and I would really appreciate some help

     

    Thanks

    file upload error

    in PHP Coding Help

    Posted

    Thanks for your reply, I missed the database name previously which I have now put in but I still get the message. I also read the link you kindly put up and to be honest being new to php I am not really any wiser as I want to put the data into an database and this explains putting it into a file?

    file upload error

    in PHP Coding Help

    Posted

    Hi I am trying o uploa some files into mysql database by keep getting the error message

     

     

    Warning: fread(): supplied argument is not a valid stream resource in C:\Program Files\xampp\htdocs\upload.php on line 4

     

    File ID: 0

     

    File Name:

     

    File Size:

     

    File Type:

     

    To upload another file Click Here

     

    I have an html form

     

    <form method="post" action="upload.php" enctype="multipart/form-data">
Description:<br>
<input type="text" name="form_description" size="40">
<input type="hidden" name="MAX_FILE_SIZE" value="1000000000">
<br>File to upload:<br>
<input type="file" name="form_data" size="40">
<p><input type="submit" name="submit" value="submit">
</form>

     

    and a php handling script

     

    <?php
mysql_connect("localhost","root","");
mysql_select_db("database_name");
$data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));
$result=MYSQL_QUERY("INSERT INTO uploads (description, data,filename,filesize,filetype) ". "VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");
$id= mysql_insert_id();
print "<p>File ID: <b>$id</b><br>";
print "<p>File Name: <b>$form_data_name</b><br>";
print "<p>File Size: <b>$form_data_size</b><br>";
print "<p>File Type: <b>$form_data_type</b><p>";
print "To upload another file <a href=http://localhost/upload.html> Click Here</a>";
?>

     

    I have been through the code and seached the forum but can't find a resolution so I would really appreciate some help!

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