justAnoob Posted April 2, 2009 Share Posted April 2, 2009 Still can't get image to display,,, just the image path. <?php // "<img src='/userimages/cars/>" this is the image path where the pics are echo "<tr><td>"; echo $row['imgpath']; //this only echos the actuall image path, not the pic ?> <?php echo "<img src='/userimages/cars/pic.jpg>"; //if i enter one of the picture names in the code, the same picture is for every entry displayed, so atleast i know that the path is correct and the images will display. ?> this might help showing you the whole script. <?php mysql_connect("xxxxxx","xxxxx","xxxxx"); mysql_select_db("xxxxxx"); $sql = mysql_query("SELECT imgpath, item_name, description, in_return FROM xxxxxx"); echo "<table border='0' CELLPADDING=5 STYLE='font-size:16px'>"; echo "<table border=1> <tr> <td><H3>Image </h3></td> <td><H3>Item Name</H3></td> <td><H3>Description</H3></td><td><H3>Seeking</H3></td></tr>"; while ($row = mysql_fetch_array($sql)) { echo "<tr><td>"; echo $row['imgpath']; echo "</td><td>"; echo $row['item_name']; echo "</td><td>"; echo $row['description']; echo "</td><td>"; echo $row['in_return']; echo "</td></tr>"; } echo "</table>"; ?> so it has to be something minor to get the images to display for each row of the datase Quote Link to comment Share on other sites More sharing options...
jonsjava Posted April 2, 2009 Share Posted April 2, 2009 <?php mysql_connect("xxxxxx","xxxxx","xxxxx"); mysql_select_db("xxxxxx"); $sql = mysql_query("SELECT imgpath, item_name, description, in_return FROM xxxxxx"); echo "<table border='0' CELLPADDING=5 STYLE='font-size:16px'>"; echo "<table border=1> <tr> <td><H3>Image </h3></td> <td><H3>Item Name</H3></td> <td><H3>Description</H3></td><td><H3>Seeking</H3></td></tr>"; while ($row = mysql_fetch_array($sql)) { echo "<tr><td>"; echo "<img src=\"".$row['imgpath']."\">"; echo "</td><td>"; echo $row['item_name']; echo "</td><td>"; echo $row['description']; echo "</td><td>"; echo $row['in_return']; echo "</td></tr>"; } echo "</table>"; ?> Quote Link to comment Share on other sites More sharing options...
thebadbad Posted April 2, 2009 Share Posted April 2, 2009 Your HTML is malformed (missing quote around image path). If $row['imgpath'] holds the image path, i.e. /userimages/cars/pic.jpg, this should work: while loop excerpt <?php while ($row = mysql_fetch_array($sql)) { echo "<tr><td>"; echo '<img src="' . $row['imgpath'] . '" alt="" />'; echo "</td><td>"; echo $row['item_name']; echo "</td><td>"; echo $row['description']; echo "</td><td>"; echo $row['in_return']; echo "</td></tr>"; } ?> Edit: Got beaten Quote Link to comment Share on other sites More sharing options...
justAnoob Posted April 2, 2009 Author Share Posted April 2, 2009 thanks everyone. Quote Link to comment Share on other sites More sharing options...
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