Hello,
I try this: <?= base_url($image['src']; ?>
PHP Code:
<br><br>
<table style="padding: 10px;">
<tr>
<td><b>EDIT CAPTION</b></td>
<td><center><b>IMAGES</b></td>
<td><b>DELETE</b></center></td>
</tr>
<tr>
<td><input type="text" name="caption" value=""></td>
<td><img src="<?php echo base_url('uploads/'.<?= base_url($image['src']; ?>'); ?>" height="300" width="400"></td>
<td><button class="delete">DELETE</button></td>
</tr>
<tr>
<td><input type="text" name="caption" value=""></td>
<td><img src="<?php echo base_url('uploads/pic2.jpg'); ?>" height="300" width="400"></td>
<td><button class="delete">DELETE</button></td>
</tr>
</table>
</div>
<?php echo form_close(); ?>
<?php if (isset($success_msg)) { echo $success_msg; } ?>
</div>
And it does not read my codes.
Parse error: syntax error, unexpected '<' in
C:\Program Files (x86)\EasyPHP-Devserver-16.1\eds-www\CompanyGiondaCI\application\views\addslideshows.php on line
112
A PHP Error was encountered
Severity: Parsing Error
Message: syntax error, unexpected '<'
Filename: views/addslideshows.php
Line Number: 112
Backtrace:
I revise it into this:
Line 112: is <td><img src="<?php echo base_url('uploads/'.<?= base_url($image['src']; ?>); ?>" height="300" width="400"></td>
PHP Code:
<br><br>
<table style="padding: 10px;">
<tr>
<td><b>EDIT CAPTION</b></td>
<td><center><b>IMAGES</b></td>
<td><b>DELETE</b></center></td>
</tr>
<tr>
<td><input type="text" name="caption" value=""></td>
<td><img src="<?php echo base_url('uploads/'.<?= base_url($image['src']; ?>); ?>" height="300" width="400"></td>
<td><button class="delete">DELETE</button></td>
</tr>
<tr>
<td><input type="text" name="caption" value=""></td>
<td><img src="<?php echo base_url('uploads/pic2.jpg'); ?>" height="300" width="400"></td>
<td><button class="delete">DELETE</button></td>
</tr>
</table>
</div>
<?php echo form_close(); ?>
<?php if (isset($success_msg)) { echo $success_msg; } ?>
</div>
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