PHP reading database with a variable

Question

I want to read 5 attributes of the database. The 5 attributes have the names post_image_1, post_image_2, post_image_3, post_image_4 and post_image_5. Now I want to show the 5 images on my page, with a for loop.

Here is the loop:

for($i = 1; $i <= 5; $i++){
    echo "<img src='image/$row[post_image_$i].png' height='250px' width='250px'>";
}

Now I get an error:

Parse error: syntax error, unexpected '$i' (T_VARIABLE), expecting ']' in

I hope that's enough info to help me. :P

You are trying to mix a constant and variable name together in your array index key as far as PHP is concerned. — Crackertastic
– Crackertastic, Commented Feb 6, 2015 at 21:18

rm-vanda · Accepted Answer · 2015-02-06 21:18:46Z

4

Since you're using arrays, do this:

for($i = 1; $i <= 5; $i++){
    echo '<img src="image/'.$row['post_image_'.$i].'png" height="250px" width="250px">';
}

answered Feb 6, 2015 at 21:18

rm-vanda

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2 Comments

LG_PDX Over a year ago

beat me by seconds... :)

rm-vanda Over a year ago

A for effort ;) - in fact, if OP catches this, you can take the answer points - the upvotes already got me back over 2,000 -

LG_PDX · Accepted Answer · 2015-02-06 21:19:55Z

2

Try this, using string concatenation, it gets at the right field in the array. I'm assuming $row is declared already and contains the keys 'post_image_1', 'post_image_2'...

for($i = 1; $i <= 5; $i++){
    echo "<img src='image/" . $row['post_image_' . $i] . ".png' height='250px' width='250px'>";
}

answered Feb 6, 2015 at 21:19

LG_PDX

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