OriginalSunny Posted March 8, 2006 Share Posted March 8, 2006 Hi,I am trying to get a picture as my link to another page in php. The thing is when the user will click on the picture a query should also be performed. I am currently using the following code for the user to select an option and the queryto be performed:[u]FIRST PAGE[/u][i] foreach($food_categories as $key => $subarray) { echo "<h3>$key</h3>"; echo "<ul>"; foreach($subarray as $type) { echo "<input type='radio' name='interest' value='$type'><b>$type</b><br>\n"; } echo "</ul>"; } echo "<p><input type='submit' name='Products' value='Select Category'>\n [/i][u]SECOND PAGE[/u][i] $sql_cat = "SELECT DISTINCT class FROM food ORDER BY class"; $result = mysql_query($sql_cat,$connect) or die("sql_cat: ".mysql_error($connect)); while($row = mysql_fetch_array($result)) { $food_categories[$row['type']][]=$row['class']; } include("catalogContract.inc");[/i]As you can see a list of the class variables is displayed when the user clicks on the select category button. Now how do i use a picture instead?? I have stored the picture in my food database as foodpic.Thanks. Quote Link to comment Share on other sites More sharing options...
littlened Posted March 9, 2006 Share Posted March 9, 2006 [!--quoteo(post=352860:date=Mar 8 2006, 02:32 PM:name=OriginalSunny)--][div class=\'quotetop\']QUOTE(OriginalSunny @ Mar 8 2006, 02:32 PM) [snapback]352860[/snapback][/div][div class=\'quotemain\'][!--quotec--]Hi,I am trying to get a picture as my link to another page in php. The thing is when the user will click on the picture a query should also be performed. I am currently using the following code for the user to select an option and the queryto be performed:[u]FIRST PAGE[/u][i] foreach($food_categories as $key => $subarray) { echo "<h3>$key</h3>"; echo "<ul>"; foreach($subarray as $type) { echo "<input type='radio' name='interest' value='$type'><b>$type</b><br>\n"; } echo "</ul>"; } echo "<p><input type='submit' name='Products' value='Select Category'>\n [/i][u]SECOND PAGE[/u][i] $sql_cat = "SELECT DISTINCT class FROM food ORDER BY class"; $result = mysql_query($sql_cat,$connect) or die("sql_cat: ".mysql_error($connect)); while($row = mysql_fetch_array($result)) { $food_categories[$row['type']][]=$row['class']; } include("catalogContract.inc");[/i]As you can see a list of the class variables is displayed when the user clicks on the select category button. Now how do i use a picture instead?? I have stored the picture in my food database as foodpic.Thanks.[/quote]the simplest way to do it would be make the image a link, and use PHP do make the link dynamic.for exampleecho "<h3>$key</h3>"; echo "<ul>"; foreach($subarray as $type) { <a href="index.php?type=<?php echo $type; ?>"><img src="pic.gif" border="0"></a>"; }if you need the image to be different on each one, you could store the url of the image in a database or an array and do the same as what you do with the $type variable.eg<a href="index.php?type=<?php echo $type; ?>"><img src="<?php echo $picurl;?>" border="0"></a>"; Quote Link to comment Share on other sites More sharing options...
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