Load variables from while loop

Question

I have a MySQL table which contains page content, each line of content has a page_id number

I want to load all the variables for the page I'm on, then display them where they need to be...

I'm naming the variables: div_1, div_2, div_3 etc etc

here's what I'm trying:

PHP

$home_p1 = mysql_query("select div_content from page_content where (page_id = '1');",$db);

$i=1;
while ($p1 = mysql_fetch_array($home_p1)){
$div_.$i = $p1[0];
$i++;
}

then throughout the page i can

echo $div_1;
blah blah blah
echo $div_2
blah blah blah
echo $div_3
etc etc

But I'm getting nothing back, I'm not even sure if this is the correct logic to do such a thing?

axiomer · Accepted Answer · 2012-02-28 10:38:14Z

1

Use arrays.

$div[$i] = $p1[0];

And through the page you can:

echo $div[1];
blah blah blah
echo $div[2];
blah blah blah
echo $div[3];
etc etc

If you want to keep your structure structure then use:

$tempname = "div_".$i;
$$tempname=$p1[0];

For more details see this (first solution) and this (second solution).

edited Feb 28, 2012 at 10:38

answered Feb 28, 2012 at 10:32

axiomer

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Comments

Stefan · Accepted Answer · 2012-02-28 10:33:24Z

0

organize it in arrays like this:

$home_p1 = mysql_query("select div_content from page_content where (page_id = '1');",$db);
$results = array();
$i=1;
while ($p1 = mysql_fetch_array($home_p1)){
    //$div_.$i = $p1[0];
    $results[] = $p1[0];
    $i++;
}

then you can access the first line by

echo $results[0]; // second by $results[1] and so on.

have fun!

answered Feb 28, 2012 at 10:33

Stefan

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Comments

davidethell · Accepted Answer · 2012-02-28 10:36:03Z

0

You can't create variables like this:

$div_.$i = $p1[0];

You should put them into an array like this:

$div_array = array();

then

$div_array[$i] = $p1[0];

Now you can output them by referencing the array directly:

echo $div_array[1];

Or else you can expand the array into variables using extract:

extract($div_array, EXTR_PREFIX_ALL, 'div_');

Now you have your $div_$i variables.

answered Feb 28, 2012 at 10:36

davidethell

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Comments

Vitalii Lebediev · Accepted Answer · 2012-02-28 10:38:32Z

0

In order to accopmlish what you are trying to, you should do something like next:

$home_p1 = mysql_query("select div_content from page_content where (page_id = '1');",$db);

$i=1;
while ($p1 = mysql_fetch_array($home_p1)){
  $name = 'div_' . $i;
  $$name = $p1[0];
  $i++;
}

You will get the list of variables with names $div_1, $div_2, etc. http://www.php.net/manual/en/language.variables.variable.php

But the best way, as it was said, to use arrays.

$home_p1 = mysql_query("select div_content from page_content where (page_id = '1');",$db);

$i=1;
$divs = Array();
while ($p1 = mysql_fetch_array($home_p1)){
  $divs['div_' . $i] = $p1[0];
  $i++;
}

answered Feb 28, 2012 at 10:38

Vitalii Lebediev

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