PHP For Loop Working

Question

I have few lines of code that I think will work normally in any language. But this is not working in PHP. In my case I want to print numbers in ascending order. The code I have written is below:

$i = 0;
printf("<p>Numbers in Ascending Order : ");
for (;++i <= 10;) {
    printf("%3d", $i);
    printf("\n\n");
}

But I get a syntax error which is given below:

Parse error: syntax error, unexpected '<=' (T_IS_SMALLER_OR_EQUAL), expecting

Why is PHP displaying an error message like this ?

see manual: http://php.net/manual/en/control-structures.for.php — kamal pal, May 09 '16 at 06:22
Mayhaps if you added a `$` in front of the `++i` like so: `++$i`. — Rasclatt, May 09 '16 at 06:24

Thamilhan · Accepted Answer · 2016-05-09T06:36:03.223

2

You are missing $ in for loop variable

it should be:

$i = 0;
printf("<p>Numbers in Ascending Order : ");
for (;++$i <= 10;) {
        ^
    printf("%3d", $i);
    printf("\n\n");
}

edited May 09 '16 at 06:36

answered May 09 '16 at 06:24

Thamilhan

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3

yes perfect. I have added same in comment. :) – RJParikh May 09 '16 at 06:28

score 0 · Answer 2 · edited May 09 '16 at 06:26

0

In php all variable names start with $ character. On your loop there is plain i. Add $ sign and it will work.

edited May 09 '16 at 06:26

Rasclatt

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answered May 09 '16 at 06:24

daFool

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score 0 · Answer 3 · answered May 09 '16 at 06:26

0

You missed $ in your variable.Replace ++i with ++$i.It should be like this:

$i = 0;
printf("<p>Numbers in Ascending Order : ");
for (;++$i <= 10;) {
    printf("%3d", $i);
    printf("\n\n");
}

answered May 09 '16 at 06:26

Chonchol Mahmud

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score 0 · Answer 4 · answered May 09 '16 at 06:26

The $ sign is missing from the i variable in the loop. In PHP the $ sign goes everywhere not just in variable decleration

    $i = 0;
    printf("<p>Numbers in Ascending Order : ");
    for (;++$i <= 10;) {
       printf("%3d", $i);
       printf("\n\n");
    }

PHP For Loop Working

4 Answers4