reazik Posted March 25, 2006 Share Posted March 25, 2006 I keep getting an "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ----------- on line 13" error for this script (the line in question is red)._________<? mysql_pconnect("localhost","xxxxxxxx","xxxxxxxx"); mysql_select_db("psnincne_JobForms"); $result = mysql_query("select * from `psnincne_job_list`"); ?><? [!--coloro:#FF0000--][span style=\"color:#FF0000\"][!--/coloro--]while($r=mysql_fetch_array($result)) { [!--colorc--][/span][!--/colorc--]$job_code = $r["job_code"]; $job_title = $r["job_title"]; $job_state = $r["job_state"]; $job_date = $r["job_date"]; echo "$job_code <br> $job_title <br> $job_state <br> $job_date"; } ?>_________What's the problem? Link to comment https://forums.phpfreaks.com/topic/5740-help-another-question-from-a-nooooobie/ Share on other sites More sharing options...
AV1611 Posted March 25, 2006 Share Posted March 25, 2006 change this:[code]$result = mysql_query("select * from `psnincne_job_list`"); [/code]to this:[code]$result = mysql_query("select * from `psnincne_job_list`") or die (mysql_error()); [/code]then post any errors Link to comment https://forums.phpfreaks.com/topic/5740-help-another-question-from-a-nooooobie/#findComment-20458 Share on other sites More sharing options...
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