php concating variable and accessing array

Question

is somthing like that possible?

for ($i = 1; $i <= 15; $i++) {
        $var = "bild" . $i;
        if (!empty ($row->bild.$i)) {
            echo '
                <div class="item">
                        <img border="no" class="bildschatten" src="include/images/projekte/'.$row->bild . $i.'" />
                </div>
            ';
        }
    }

how should it look correctly? problem is:

I fetch objects from a mysql database, in this table each object has 15 images

bild1 .. bild15

Now I want to iterate over this 15 images and want to check if the colum is empty or not.

the problem I have is here:

$row->bild.$i

that contains not the value of column bild1.. bild15 ... it only contains the value of $i

thanks

mysterious tried this bevore ... its working now sorry for this stupid question — Felix
– Felix, Commented Oct 15, 2016 at 14:07

krasipenkov · Accepted Answer · 2016-10-15 14:11:31Z

1

This code should work:

$row->{'bild'.$i}

answered Oct 15, 2016 at 14:11

krasipenkov

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Comments

Murad Hasan · Accepted Answer · 2016-10-15 14:11:28Z

1

This is called variable of variables, You need to access something like this $row->$var.

Complete Solution:

for ($i = 1; $i <= 15; $i++) {
    $var = "bild" . $i;
    if (!empty ($row->$var)) {
        echo '
            <div class="item">
                    <img border="no" class="bildschatten" src="include/images/projekte/'.$row->$var.'" />
            </div>
        ';
    }
}

For more details checkout the Manual.

answered Oct 15, 2016 at 14:11

Murad Hasan

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Comments

ScaisEdge · Accepted Answer · 2016-10-15 14:14:28Z

1

You could fetch the result as an array uning

    $row  =  db_fetch_array($result)

and then accessing

    $var = "bild" . $i;
    if (!empty ($row[$var])) {

edited Oct 15, 2016 at 14:14

answered Oct 15, 2016 at 13:53

ScaisEdge

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2 Comments

Felix Over a year ago

then I got this: Fatal error: Cannot use object of type stdClass as array in the row var is this: while ($row = db_fetch_object($sql)) {

ScaisEdge Over a year ago

@Felix . like indicated at the beginning of the answer you shuold use fetch the result as array using db_fetch_array($result)

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