Jump to content

help


spearchilduser

Recommended Posts

can anyone see a error with this code ?

 

$query = mysql_query("SELECT * FROM project WHERE Module_Code ='$schemelevel' AND Available ='Yes'");

$result = mysql_query($query);

 

while ($myrow = mysql_fetch_row($result)) {

 

seems to be a issue using fetch_row ($result) telling me theres a issue with the resource

 

Link to comment
https://forums.phpfreaks.com/topic/259742-help/
Share on other sites

mysql_query expects it's first argument to be a string containing valid SQL, which is what you have contained in $query.

With $result, you are attempting to query a query resource, which is invalid, the code should read:

 

$query = "SELECT * FROM project WHERE Module_Code = '$schemelevel' AND Available = 'Yes'";
$result = mysql_query($query);
while ($myrow = mysql_fetch_row($result))
{
//code
}

Link to comment
https://forums.phpfreaks.com/topic/259742-help/#findComment-1331217
Share on other sites

Is there any other code in the page?

Try this debugging code and post the results:

 

$sql = "SELECT * FROM project WHERE Module_Code = '$schemelevel' AND Available = 'Yes'";
$result = mysql_query($sql) or die($sql .'<br />'. mysql_error());
if(mysql_num_rows($result) == 0)
{
    echo "no results";
    exit;
}
while ($myrow = mysql_fetch_row($result))
{
//code
}

Link to comment
https://forums.phpfreaks.com/topic/259742-help/#findComment-1331221
Share on other sites

alright, then try this:

 

$sql = "SELECT * FROM project WHERE Module_Code = '$schemelevel' AND Available = 'Yes'";
$result = mysql_query($sql) or die($sql .'<br />'. mysql_error());
if(mysql_num_rows($result) == 0)
{
    echo "no results <br/ >";
    echo $sql;
    exit;
}
while ($myrow = mysql_fetch_row($result))
{
//code
}

Link to comment
https://forums.phpfreaks.com/topic/259742-help/#findComment-1331225
Share on other sites

<?php
$db = mysql_connect("", "","");
mysql_select_db("");

$schemelevel = $_POST['schemelevel'];

$sql = "SELECT * FROM project WHERE Module_Code = '$schemelevel' AND Available = '1'";
$result = mysql_query($sql) or die($sql .'<br />'. mysql_error());
if(mysql_num_rows($result) == 0)
{
    echo "no results <br/ >";
    echo $sql;
    exit;
}
while ($myrow = mysql_fetch_row($result))
{
echo "<table border=1>\n";
echo "<tr><td>Module Code</td><td>Title</td><td>First_Supervisor</td><td>Level</td><td>Description</td></tr>\n";




echo "<tr><td><A HREF='http://students.comp.glam.ac.uk/08030472/project/displayproject.php?T1=$myrow[0]'>$myrow[0]</A></td><td>$myrow[1]</td><td>$myrow[2]</td><td>$myrow[3]</td><td>$myrow[4]</td></tr>\n";
	//printf("<tr><td>%s</td><td>%s</td><td>%s</td><td>%s</td><td>%s</td></tr>\n",
	//echo"<A HREF='http://students.comp.glam.ac.uk/08030472/project/displayproject.php?T1=$myrow[0]'>$myrow[0]</A>"$myrow[0],$myrow[1],$myrow[2],$myrow[3],$myrow[4]);
}
echo "</table>\n";
?>
</body>
</html>

Link to comment
https://forums.phpfreaks.com/topic/259742-help/#findComment-1331255
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.