justineaguas Posted November 12, 2009 Share Posted November 12, 2009 I am using an array to check and compare the values in one variable, If the variable exists, the set of values will be inserted in the table of the variable. I am successful in inserting the values inside the mySQL table. However, I receive an error: Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\datbas\do_addpage_off.php on line 98 while ($row=mysql_fetch_array($result, MYSQL_ASSOC) or die(mysql_error())) //error here LINE 98 { $temp=$row['BRGY_name']; if ($temp==$brgy) { include ('connect.php'); $get = "SELECT BRGY_id FROM BRGY_info WHERE BRGY_name='$brgy'"; $res = mysql_query($get) or die(mysql_error()) or die(mysql_error()); $row = mysql_fetch_array($res, MYSQL_ASSOC); $sql = "INSERT INTO OFF_info(OFF_fname,OFF_lname,OFF_position,OFF_contact,BRGY_id,OFF_status)VALUES('$fname','$lname','$position','$contact','{$row['BRGY_id']}','$num')"; $result = mysql_query($sql) or die(mysql_error()); echo " <br /><br /> <center> Thank you for adding a new barangay.<br /> <span class='linkBody'> <a href='addpage.php'>Add Another?</a> </span> <br /> <br /> </center>"; } Link to comment https://forums.phpfreaks.com/topic/181256-solved-warning-mysql_fetch_array-expects-parameter-1-to-be-resource/ Share on other sites More sharing options...
PFMaBiSmAd Posted November 12, 2009 Share Posted November 12, 2009 Your code inside of the while() loop is reusing the $result variable, so the next time the while() condition is evaluated, you have a TRUE/FALSE value instead of the original result resource. Please be careful when reusing variables. Link to comment https://forums.phpfreaks.com/topic/181256-solved-warning-mysql_fetch_array-expects-parameter-1-to-be-resource/#findComment-956206 Share on other sites More sharing options...
justineaguas Posted November 12, 2009 Author Share Posted November 12, 2009 So how do I fix this? I'm sorry I'm a newbie in PHP Link to comment https://forums.phpfreaks.com/topic/181256-solved-warning-mysql_fetch_array-expects-parameter-1-to-be-resource/#findComment-956208 Share on other sites More sharing options...
sasa Posted November 12, 2009 Share Posted November 12, 2009 change variable name $result to $resul1 in this $sql = "INSERT INTO OFF_info(OFF_fname,OFF_lname,OFF_position,OFF_contact,BRGY_id,OFF_status)VALUES('$fname','$lname','$position','$contact','{$row['BRGY_id']}','$num')"; $result = mysql_query($sql) or die(mysql_error()); Link to comment https://forums.phpfreaks.com/topic/181256-solved-warning-mysql_fetch_array-expects-parameter-1-to-be-resource/#findComment-956221 Share on other sites More sharing options...
justineaguas Posted November 12, 2009 Author Share Posted November 12, 2009 Thanks!! It fixed Link to comment https://forums.phpfreaks.com/topic/181256-solved-warning-mysql_fetch_array-expects-parameter-1-to-be-resource/#findComment-956227 Share on other sites More sharing options...
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