PHP combine values from loop

Question

If I want to display the text as follows in example, how can it be done?

$tmp=mysql_query("SELECT ... FROM ....");
$total=mysql_num_rows($tmp);
$tekst1='Your total number of friends are: '.$total.'. Your friends are: ';
while($p=mysql_fetch_array($tmp)) {
$tekst2=$p['friend'].", ";
}
$text_complete=$tekst1.$tekst2;
echo $text_complete;

$text_complete should look like: "Your total number of friends are: 3. Your friends are: John, Mike, Michael". But, using this code above I get "Your total number of friends are: 3. Your friends are: John,"

How can I combine both texts, the one obtained from the loop and the one which is fixed?

Nicola Peluchetti · Accepted Answer · 2011-05-31 15:19:53Z

1

I think you shoul do like this:

       $tmp=mysql_query("SELECT ... FROM ....");
       $total=mysql_num_rows($tmp);
       $tekst1='Your total number of friends are: '.$total.'. Your friends are: ';
       $friends = array();
       while($p = mysql_fetch_array($tmp)) {
             $friends[] = $p['friend'];
       }
       $tekst2= implode(', ', $friends);
       $text_complete=$tekst1.$tekst2;
       echo $text_complete;

otherwise you overwrite the variable each time!

EDIT - i corrected the code to use implode as in Xaerxess answer. Previousli i just concatenated and use a substr to remove trailing comma and space

edited May 31, 2011 at 15:19

answered May 31, 2011 at 13:09

Nicola Peluchetti

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4 Comments

Tim Baas Over a year ago

That's right but you should use it on $tekst1, so you don't need $tekst2 and $text_complete..

Sergio Over a year ago

@Nicola Peluchetti - I get error message: "Undefined variable: tekst2 on line 25". That line is: "$tekst2.=$p['friend'].", ";"

Sergio Over a year ago

@Nicola Peluchetti -I found why this happens. Now it's working fine. Thanks.

keymone Over a year ago

implode version in other comments is better

George Cummins · Accepted Answer · 2011-05-31 13:11:01Z

1

You can use:

$tekst2 .= $p['friend'].", ";

To add to the variable (rather than overwriting it. Then remove the trailing comma:

$text_complete=$tekst1. rtrim( $tekst2, "," );

answered May 31, 2011 at 13:11

George Cummins

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Comments

Billy Moon · Accepted Answer · 2011-05-31 13:11:56Z

1

$tekst2 = new Array();
while($p=mysql_fetch_array($tmp)) {
    $tekst2[] = $p['friend'];
}
$text_complete = $tekst1 . implode(', ',$tekst2);

But better would be to use GROUP_CONCAT in sql

answered May 31, 2011 at 13:11

Billy Moon

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George Cummins · Accepted Answer · 2011-05-31 13:13:47Z

1

$tmp=mysql_query("SELECT ... FROM ....");
$total=mysql_num_rows($tmp);
$tekst = '';
$tekst='Your total number of friends are: '.$total.'. Your friends are: ';
while($p=mysql_fetch_array($tmp)) {
$tekst.=$p['friend'].", ";
}
echo $tekst;

edited May 31, 2011 at 13:13

George Cummins

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answered May 31, 2011 at 13:11

Ruben

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Comments

Itai Sagi · Accepted Answer · 2011-05-31 13:18:48Z

1

        $tmp=mysql_query("SELECT ... FROM ....");
        $total=mysql_num_rows($tmp);
        $tekst1='Your total number of friends are: '.$total.'. Your friends are: ';
        while($p=mysql_fetch_array($tmp)) 
           {
           $tekst1 .= $p['friend'].", "; // add friend names from MySQL result.
           }
        $tekst1 = substr($tekst2, 0, -2);//Remove comma.
        echo $tekst1; // Outputs the wanted string

answered May 31, 2011 at 13:18

Itai Sagi

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Comments

Grzegorz Rożniecki · Accepted Answer · 2011-05-31 14:59:09Z

1

Use PHP implode / join funciton, so there isn't any trailing comma at the end. In this case:

$fiends = array();
while($p = mysql_fetch_array($tmp)) {
     $friends[] = $p['friend'];
}
implode(', ', $friends);

EDIT: I basically doubled Billy Moon answer. What's more don't use mysql_* functions - they're ~~deprecated~~ not recommended - see comments below.

edited May 31, 2011 at 14:59

answered May 31, 2011 at 13:13

Grzegorz Rożniecki

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4 Comments

George Cummins Over a year ago

mysql_* functions are deprecated by whom? Reference?

Grzegorz Rożniecki Over a year ago

@George Cummins: According to PHP docs mysql_* extention isn't developed, isn't recommended by MySQL for new projects, doesn't support all MySQL 4.1+ functionality. As it says: If you are using MySQL versions 4.1.3 or later it is strongly recommended that you use the mysqli extension instead.

George Cummins Over a year ago

There is a difference between the details you indicated and deprecation. Deprecation is "a status applied to software features to indicate that they should be avoided." (en.wikipedia.org/wiki/Deprecation) The mysql_* functions need not be avoided in suitable circumstances. PDO or mysqli* may be more suitable in some cases, but not all.

Grzegorz Rożniecki Over a year ago

@George Cummins: True, I've used wrong word. I've already edited my answer, thanks!

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