Displaying a single image and data

[mike6256]

Hi all
I am extremely new to php and need some help

I found a code that Atli addressed in a 11/23/07 thread and have been working with it-all is great. My main question is how do I search for a single file and display the image for that single file. The code example by Atli will list all files

Thanks

[Dormilich]

which of the 3 threads do you mean? (a link would be really helpful)

[mike6256]

Sorry-here is the link

http://bytes.com/topic/php/insights/740327-uploading-files-into-mysql-database-using-php

[Markus]

Check out MySQL's WHERE clause.

[mike6256]

Thanks for the WHERE clause link--I gave it a shot but get the result "There are no files in database" any advice for a rookie?
# Query for existing files
$result = mysqli_query($d bLink, "SELECT FileID, FileName, FileMime, FileSize, ToolNumber Created FROM FileStorage Where Toolnumber = 'textfield' ");

[Dormilich]

well, obviously there is no Toolnumber field with the value of 'textfield'.

besides, make sure, all field names are correctly written (SQL can be/is case sensitive)

[mike6256]

Thanks for the rapid reply
I did find That Toolnumber should be ToolNumber-corrected it but still the same
will 'textfield' pull over from my search page?

[hoopy]

Isn't textfield meant to be $textfield i.e a variable?

[mike6256]

I tried it with the $ variable but get a undefined variable error

[hoopy]

Lets see your code, you need to take whatever the user entered in the text box, put it into a variable then use that in the SQL to retrieve.

[mike6256]

here is the code I am using that gives me undefined variable error

<?php 
# Connect to the database 
$dbLink = mysqli_connect("xx.xx.xx.xx", "xxxxx", "xxxxxxxx", "tools"); 
if(mysqli_connect_errno()) { 
    die("MySQL connection failed: ". mysqli_connect_error()); 
} 
  
# Query for a list of all existing files 
$result = mysqli_query($dbLink, "SELECT FileID, FileName, FileMime, FileSize, ToolNumber Created FROM FileStorage Where ToolNumber = [$textfield]"); 
  
# Check if it was successfull 
if($result)  
{ 
  
    # Make sure there are some files in there 
    if(mysqli_num_rows($result) == 0) { 
        echo "<p>There are no files in the database</p>"; 
    } 
    else 
    { 
        # Print the top of a table 
        echo "<table width='100%'><tr>"; 
        echo "<td><b>Name</b></td>"; 
        echo "<td><b>Mime</b></td>"; 
        echo "<td><b>Size (bytes)</b></td>"; 
        echo "<td><b>Created</b></td>"; 
        echo "<td><b>Tool Number</b></td>";
        echo "<td><b>&nbsp;</b></td>"; 	
        echo "</tr>"; 
  
        # Print each file 
        while($row = mysqli_fetch_assoc($result)) 
        { 
            # Print file info 
            echo "<tr><td>". $row['FileName']. "</td>"; 
            echo "<td>". $row['FileMime']. "</td>"; 
            echo "<td>". $row['FileSize']. "</td>"; 
            echo "<td>". $row['Created']. "</td>"; 
            echo "<td>". $row['ToolNumber']. "</td>";
  
            # Print download link 
            echo "<td><a href='get_file.php?id=". $row['FileID'] ."'>Download</a></td>"; 
            echo "</tr>"; 
        } 
  
        # Close table 
        echo "</table>"; 
    } 
  
    # Free the result 
    mysqli_free_result($result); 
} 
else  
{ 
    echo "Error! SQL query failed:"; 
    echo "<pre>". $dbLink->error ."</pre>"; 
} 
  
# Close the mysql connection 
mysqli_close($dbLink); 
?>

[Dormilich]

I'm not sure what the square brackets are for... of cause $textfield has to be defined somewhere (looks like some form data).

// just one possibility...
$textfield = $_POST['user_input'];

$result = mysqli_query($dbLink, "SELECT FileID, FileName, FileMime, FileSize, `ToolNumber Created` FROM FileStorage WHERE ToolNumber = " . mysql_real_escape_string($textfield));
// use ' around the value if it is a string

[mike6256]

would something like this work?

$textfield = $_POST['textfield']; 
  
$result = mysqli_query($dbLink, "SELECT FileID, FileName, FileMime, FileSize, ToolNumber, Created FROM FileStorage WHERE ToolNumber = " . mysql_real_escape_string($textfield));

here is the code for the search

<form id="form1" name="form1" method="get" action="list_files.php">
  <label>Tool Number
  <input type="text" name="textfield" />
  </label>
  <p>
    <label>
    <input type="submit" name="Submit" value="Submit" />
    </label>
  </p>

sorry for so many questions. I have only been doing this since Monday

[hoopy]

Your method is get on the form, not post so change this:

<form id="form1" name="form1" method="get" action="list_files.php">

To:

<form id="form1" name="form1" method="post" action="list_files.php">

And that should work OK.

Or alternatively change:

$textfield = $_POST['textfield'];

To:

$textfield = $_GET['textfield'];

Cheers

[mike6256]

Well I did the reccommended changes and now get another error:

Error! SQL query failed:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
PHP Warning: mysql_real_esca pe_string() [function.mysql-real-escape-string]: Access denied for user 'ODBC'@'localho st' (using password: NO) in C:\Inetpub\wwwr oot\tlsc\list_f iles.php on line 12 PHP Warning: mysql_real_esca pe_string() [function.mysql-real-escape-string]: A link to the server could not be established in C:\Inetpub\wwwr oot\tlsc\list_f iles.php on line 12

This is line 12:

$result = mysqli_query($dbLink, "SELECT FileID, FileName, FileMime, FileSize, ToolNumber, Created FROM FileStorage WHERE ToolNumber = " . mysql_real_escape_string($textfield));