Loop Issue Counting Year In PHP

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陌清茗
陌清茗 2021-01-22 10:35

I have created a loop which will display date 2004 to 2014 in a formatted way. But the problem is, it is showing 204 instead of 2004 and continue this till 209.. So, how to show

10条回答
  •  一个人的身影
    2021-01-22 11:14

    According to your code, you can try this. Though its not a standard way:

    ";
            $ax++;
        }
        $ax = 1;
        while ($ax <= 31) {
            echo "$ax Feb 200$yar 
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Mar 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Apr 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax May 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Jun 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Jul 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Aug 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Sep 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Oct 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Nov 200$yar
    "; $ax++; } $ax = 1; while ($ax <= 31) { echo "$ax Dec 200$yar
    "; $ax++; } $yar++; } $yr = 10; while ($yr <= 14) { $x = 1; while ($x <= 31) { echo "$x Jan 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Feb 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Mar 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Apr 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x May 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Jun 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Jul 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Aug 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Sep 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Oct 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Nov 20$yr
    "; $x++; } $x = 1; while ($x <= 31) { echo "$x Dec 20$yr
    "; $x++; } $yr++; } ?>

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