visitor (0 QPoints)
  • FR
  • EN
  • NL
  • DE
  • ES
315 experts, 1193 registered users, 1659 questions already answered
European Experts Exchange, the very best site for high-quality IT solutions

New Improved Search!

 


05/10/2011 1h30 : Steve Jobs is dead, the father of Apple ][ is gone, we are all orphaned.

Languages :: PHP :: can't increment
By: pdd U.S.A.  Date: 25/09/2003 00:00:00  English  Points: 75 Status: Answered
Quality : Excellent
Hi !

ok, i have this code

<?php

while ($nume = mysql_fetch_array($sql_result_numele_p_d)){

$i=0;

echo ("<option value=\"$i\">$nume[$i]</option>\n");

$i++;

}

?>

mysql_fetch_array($sql_result_numele_p_d) returns an array with 18 elements

now my question. why is the value of every <option> tag 0 and how do i increment that whit every new <option> tag
By: Michael701 Date: 25/09/2003 04:16:00 English  Type : Assist
first your $i=0 is INSIDE the while loop, it will reset each time;
do you really want this? or are you trying to use one field from each row as the option?


<?php
$nume = mysql_fetch_array($sql_result_numele_p_d);
$i=0;

foreach ($nume as $value)
{
echo ("<option value=\"$i\">$nume[$i]</option>\n");
$i++;
}
?>
By: VGR Date: 25/09/2003 04:29:00 English  Type : Comment
naaaaa do this :

<?php

$i=0;
while ($nume = mysql_fetch_array($sql_result_numele_p_d)){
echo "<option value=\"$i\">{$nume[$i]}</option>\n";
$i++;
} // while records

?>


PS : NEVER forget to write things like "{$nume[$i]}" when printing array elements in strings :D
By: VGR Date: 25/09/2003 04:31:00 English  Type : Answer
ah sorry, I misread your problem. What a strange idea to access by value of index an associative array when you used mysql_fetch_array()... You could as well have used a non_assoc mysql_* function

<?php

while ($nume = mysql_fetch_array($sql_result_numele_p_d)) for ($i=0;$i<18:$i++) echo "<option value=\"$i\">{$nume[$i]}</option>\n";

?>
By: Michael701 Date: 25/09/2003 04:38:00 English  Type : Comment
VGR, i'm not sure if he wants 18 values from each record or 1 value from 18 records.

moving the $i=0; outside the loop will give him first field from first record, second field from second record.... (very odd)
By: VGR Date: 25/09/2003 17:20:00 English  Type : Comment
he's doing a while (mysql_fetch_array()) and states that "mysql_fetch_array($sql_result_numele_p_d) returns an array with 18 elements"

so I bet for the first solution : X records of 18 elements
By: PHPaul Date: 25/09/2003 17:56:00 English  Type : Comment
VGR, bit off topic, but I've never done this...
echo "<option value=\"$i\">{$nume[$i]}</option>\n";

I'm refering to the {$nume[$i]} piece.
Shouldn't you use echo " bla " . $nume[$i] . "bla"; instead?

thx
By: VGR Date: 25/09/2003 18:05:00 English  Type : Comment
no

both are accepted

I'm just fed up with quoting in and out, so I finally found that PHP offered a simple & elegant way to include also array elements in strings

see php doc

BTW, just try, you'll see what works, and what is shorter to type & easier to use :D
By: VGR Date: 25/09/2003 18:08:00 English  Type : Comment
<A HREF="http://be.php.net/echo">http://be.php.net/echo</a>


echo

(PHP 3, PHP 4 )
echo -- Output one or more strings
Description
[...]
Example 1. echo() examples
[...]
echo "this is {$bar['value']} !"; // this is foo !
By: VGR Date: 25/09/2003 18:09:00 English  Type : Comment
and I thus escape the escaping problem :D
By: pdd Date: 26/09/2003 19:04:00 English  Type : Comment
<?php

$i=0;

while ($nume = mysql_fetch_array($sql_result_numele_p_d)){

$i++;

echo ("<option value=\"$i\">$nume[0]</option>\n");

}

?>

ok, so this what i wanted
but why i have to use $nume[0] ?
By: VGR Date: 26/09/2003 19:44:00 English  Type : Comment
because it's NOT what you wanted !!!

moreover, the echo() line is badly written and I said it multiple times already (array references inside strings)
By: VGR Date: 26/09/2003 19:45:00 English  Type : Comment
and where is your "array of 18 elements" used ? :D :D
By: PHPaul Date: 26/09/2003 20:27:00 English  Type : Comment
lolz, weird way of doing things...

$i++; normally comes at the end of the while block.

I agree with VGR, I don't even do this:
echo "foo = $foo";
I do this...
echo "foo = " . $foo;

This should always be done with arrays, no matter what!!
eg:
echo "foo = " . $foo['some_var'];
or...
echo "foo = {$foo['some_var']}"; as VGR mentioned

--PHPaul
By: VGR Date: 28/09/2003 07:07:00 English  Type : Comment
well, don't be too hard on yourself...

The easiest thing is to write :

echo <<<EOT
here HTML with PHP $variables without any $quoting problem
Only array elements like {$_POST['toto']} need brackets
Apparently, normal array elemets like $thearray[0] will work without {}, althought I wouldn't dare to do it
best $programming ever
EOT;

Do register to be able to answer

©2010 These pages are served without commercial sponsorship. (No popup ads, etc...). Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE.
Please DO link to this page! Post to Twitter Post to Plurk Post to Yahoo Buzz Post to Delicious Post to Digg Post to Facebook Post to MySpace Post to Ping.fm Post to Reddit Post to StumbleUpon

EContact
browser fav
page generated in 327.261920 milliseconds

Why Google AdSense ads ?
compteur
 Ranking-Hits PageRank for this page