Make SQL column using PHP explode into variable

Question

Hey guys I'm having a bit of trouble trying to make my SQL column a variable in PHP.

So my SQL row contains image URL's and are formatted like this:

http://www.website.com/image.jpg,http://www.website.com/image2.jpg,http://www.website.com/image3.jpg

So now I need to display the first URL in the row.

I have this short line of code:

file = '.(explode(',', $cardata["PictureRefs"])[0]).';

I am basically trying to give the variable a value of:

$file = "http://www.website.com/image.jpg"

My current code is obviously wrong however I feel like I must be quite close. How can I achieve this?

Ohgodwhy · Accepted Answer · 2014-11-02 21:20:59Z

3

Your short hand is correct, but the string encapsulation is simply not required.

$file = explode(',', $cardata['pictureRefs'])[0]; // first image SRC

Then you can utilize it like:

echo '<img src="'.$file.'"/>';

edited Nov 2, 2014 at 21:20

answered Nov 2, 2014 at 20:54

Ohgodwhy

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2 Comments

Peter Hardy Over a year ago

I've just tried this and I'm getting this syntax error: Parse error: syntax error, unexpected '$cardata' (T_VARIABLE) in... Any idea why?

Ohgodwhy Over a year ago

@PeterHardy Missed a comma after ','. Check again, sorry.

Rogerio Coelho · Accepted Answer · 2014-11-02 22:53:19Z

0

No problem.

$file=explode(',',$cardata)[0];

answered Nov 2, 2014 at 22:53

Rogerio Coelho

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Comments

icktoofay · Accepted Answer · 2014-11-02 21:15:01Z

-1

Try this:

$file = explode(',',$cardata);
$img = $file[0];

edited Nov 2, 2014 at 21:15

icktoofay

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answered Nov 2, 2014 at 20:53

Rogerio Coelho

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1 Comment

icktoofay Over a year ago

No need to introduce that extra variable.

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