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printf and display of image


stressedsue

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Hello,

I want to display an image dynamically within the printf() function, where the image name resides on a database.  How do I embed the image name within the <img src> tag, which is within the printf() function?

Example of code that does not work:

printf("<tr>
    <td>
    <p><img src='../art/'>%s</p>
    </td>
    </tr>\n", $myrow["photoname"]");

Any help is greatly appreciated!!!
 
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https://forums.phpfreaks.com/topic/22555-printf-and-display-of-image/
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It is the last " causing you trouble (the one right before the closing parenthesis). Lose it and you should be ok.
Just curious but why use print_r for this? I think a faster solution would be something like:
[code]echo '<tr><td><p><img src="../art/' . $myrow[photoname] . '"></p></td></tr>';[/code]

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