Member Avatar for whiteyoh

Hi All,

Im getting and error from the following mysql statement when running it in the browser. Any help would be appreciated.

"you have an error in your sql syntax near -nextq accounts" on line 1"

this is the code

<?php $result = mysql_query("SELECT DISTINCT(source) FROM accounts WHERE source NOT IN ('')") or die(mysql_error());
        				    while($row = mysql_fetch_array($result)){ ?>
								 <option<?php if(!empty($_REQUEST['source']) && $_REQUEST['source']==$row['source']) echo " selected"; ?>><?=$row['source'];?>
								 <? } ?>
                                 <? $result = mysql_query("SELECT DISTINCT(original_source) FROM web146-nextq.accounts") or die(mysql_error());
        				    while($row = mysql_fetch_array($result)){ ?>
								 <option value="<? echo "3 Month Survey - " . $row['original_source'];?>"<? if(!empty($_REQUEST['source']) && $_REQUEST['source']== "3 Month Survey - " . $row['original_source']) echo " selected"; ?>>3 Month Survey - <?=$row['original_source'];?>

								 <? } ?>
                                 <? $result = mysql_query("SELECT DISTINCT(original_source) FROM web146-next6.accounts") or die(mysql_error());
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  • Latest Post Latest Post by fobos
Member Avatar for fobos

Instead of using $result multiple times, try using $result, $result2, $result and seee if that works

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