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I try to put image size in image src tag, but it still does not work. The image can be displayed fine.
<?php $files = glob("uploads/*.*"); echo "<table border =\"1\" style='border-collapse: collapse'>"; for ($row=1; $row <= 4; $row++) { echo "<tr> \n"; for ($col=1; $col<=4; $col++) { $f=$f+1; $getfile = $files[$f]; echo "<td>"; echo "<img src=$getfile > "; echo "</td>"; } echo "</tr>"; } echo "</table>";Nigel Ren
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asked Jun 12, 2018 at 6:13
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You can use getimagesize() function //php.net/manual/en/function.getimagesize.php to fetch image size
<?php $files = glob("uploads/*.*"); echo "<table border =\"1\" style='border-collapse: collapse'>"; for ($row=1; $row <= 4; $row++) { echo "<tr> \n"; for ($col=1; $col<=4; $col++) { $f=$f+1; $getfile = $files[$f]; $size = getimagesize($getfile); echo "<td>"; echo "<img src=$getfile $size[3]> "; echo "</td>"; } echo "</tr>"; } echo "</table>"; ?>answered Jun 12, 2018 at 6:20
Bhumi ShahBhumi Shah
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I am echoing an image which works fine, but then i want to set how big it needs to be, and I just can't seem to get the syntax right.
This is what I have:
echo "<img src='" . $row['imglink'] . "' height="130" width="150"> ";What am I doing wrong? If I remove the height and width it works fine.
cyborg86pl
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asked Apr 22, 2014 at 15:43
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You are mixing single and double quotes. So you get a syntax error.
This works:
echo "<img src='" . $row['imglink'] . "' height='130' width='150'> ";If you like to output double quotes you have to escape:
echo "<img src=\"" . $row['imglink'] . "\" height=\"130\" width=\"150\"> ";Or you switch to cover the string in single quotes:
echo '<img src="' . $row['imglink'] . '" height="130" width="150"> ';answered Apr 22, 2014 at 15:46
theHackertheHacker
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try changing double quotes into single quotes for these attributes
height='130' width='150'since double quotes are actually delimiting the PHP string. Optionally you may instead escape the double quotes like so
height=\"130\" width=\"150\"but this syntax is less readable and more error prone (not recommended)
answered Apr 22, 2014 at 15:45
Fabrizio CalderanFabrizio Calderan
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ok its here
echo '<img src="'$row['imglink']'" height="130" width="150">';answered Apr 22, 2014 at 15:45
4
You can uses variables inside double quotes, like this:
<? $img_url = $row['imglink']; echo "<img src='$img_url' height='130' width='150'>"; ?>answered Apr 22, 2014 at 15:50
Pedro LobitoPedro Lobito
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Sometimes the way that you're using causes to errors if you don't do it properly.Therefore you can try this way also.
<img src="<?php echo $row['imglink']?>" height="130" width="150" />if you're using this inside a loop,you can do like this.
<?php foreach($result as $rows):?> <img src="<?php echo $row['imglink']?>" height="130" width="150" /> <?php endforeach; ?>Hope this helped to you.
answered Apr 22, 2014 at 15:52
CodeCanyonCodeCanyon
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