sac0o01 Posted March 5, 2013 Share Posted March 5, 2013 I have a script that pulls urls from a database and displays the corresponding image by means of a foreach loop. I am trying to add a dropdown form to each iteration but when I run the from it only sends the values from the last iteration in the loop. The image part of the loop works perfectly. Any suggestions on a better way or how to make this one work? (excuse the mess...) <?php foreach ($image as $image_arr){?> <center> <tr> <br /><td width="70"><img src="<?php echo $image_arr;?>" alt="rotation" /><br /> </td> URL:<?php echo $image_arr;?> <br /> <?php //get ratings from database $ratingresult = mysql_query("SELECT rating FROM images WHERE url = '$image_arr' "); $ratingrow = mysql_fetch_row($ratingresult); $rating = $ratingrow[0] ?> Rating:<?php echo $rating;?></td><br /> <form name="rateform" action="rate.php" method="POST"> <input type="hidden" name="seturl" VALUE="<?php echo $image_arr ?>"/> <select name="ratedrop"> <option value="1">1 star</option> <option value="2">2 stars</option> <option value="3">3 stars</option> <option value="4">4 stars</option> <option value="5">5 stars</option> </select> <input type="submit" VALUE="rate" /> <hr /> </tr></center> <?php }?> Quote Link to comment https://forums.phpfreaks.com/topic/275251-help-using-a-dropdown-form-within-foreach-loop/ Share on other sites More sharing options...
teynon Posted March 5, 2013 Share Posted March 5, 2013 (edited) The wording of your question is a little confusing, but if I understand what you are asking, change it to this: <select name="ratedrop[]"> Edited March 5, 2013 by teynon Quote Link to comment https://forums.phpfreaks.com/topic/275251-help-using-a-dropdown-form-within-foreach-loop/#findComment-1416616 Share on other sites More sharing options...
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