Hi! I need help please! I have been thrown into having to manage a database/website and have NO idea what I am doing. I'm am a marketing/communications person and have done my best to get this up and running. (Bad split with the original programmer.) I am trying to decipher this error to know where to fix the problem.
Warning: Mysql_Fetch_Arr ay (): supplied argument is not a valid mysql result resource in/Home/Conent/A/C/C/Acce1765/Html/Reports/Boardinfluentia l.php on line 71
Same error for line 85. Both lines are the ones that start with the word while--at least that's how it appears in Notepad++.
Any help would be GREATLY appreciated!!
Thanks!
Warning: Mysql_Fetch_Arr ay (): supplied argument is not a valid mysql result resource in/Home/Conent/A/C/C/Acce1765/Html/Reports/Boardinfluentia l.php on line 71
Same error for line 85. Both lines are the ones that start with the word while--at least that's how it appears in Notepad++.
Code:
<li>Board Influential Summary
<ul>
<?php
$thequeryz = mysql_query("SELECT boardinfluentialtext FROM study_data WHERE clientchoice = '$clientx'");
while ($row = mysql_fetch_array($thequeryz)) {
$theinfo = $row['boardinfluentialtext'];
if ($theinfo == '' ) {}else{
echo '<li>'.$theinfo.'</li>';
}
} ?>
</ul>
</li></ul>
<ul>
<li>Board Influential Comments
<ul>
<?php
$thequeryz = mysql_query("SELECT boardinfluentialtext1 FROM study_data WHERE clientchoice = '$clientx'");
while ($row = mysql_fetch_array($thequeryz)) {
$theinfo1 = $row['boardinfluentialtext1'];
if ($theinfo1 == '' ) {}else{
echo '<li>'.$theinfo1.'</li>';
}
} ?>
</ul>
</li></ul>
</div>
<div id="table-2">
Any help would be GREATLY appreciated!!
Thanks!
Comment