PHP loop is not throwing the correct link

Question

for my issue i'll try to be as brief as possible

what am trying to do is reference a page with a specific id in the HTML anchor link with PHP as the below

<body>
<?php  $linkName = "Second Page"; ?>
<?php $id = 5; ?>

<a href="secondPage.php?id=<?php echo $id;?>"><?php echo $linkName?></a><br>

and it works fine

now what am trying to do is to make the $id part more dynamic by making looping the number from 1 to 10 and also providing 10 links

the code is

</head>
<body>
    <?php  $linkName = "Second Page"; ?>
    <?php $id = 5; ?>

    <?php  
        for ($i=0; $i < 10 ; $i++) { 
            echo "<a href='secondPage.php?id=<?php echo $i;?'>Link1</a>";
        };

    ?>
</body>

however what i did notice as the below images indicates when i hover on the links i noticed that i refer to a strange link

and when i cliched on it it takes me to the following link with an id that i did not want as below

http://localhost/PHP_Course/secondPage.php?id=%3C?php%20echo%201;?

i tried researching the subject and i tried escaping the quotation but it does not seem to resolve the problem

Any help please ??

You need to go learn some syntax basics. http://php.net/manual/en/language.types.string.php#language.types.string.parsing, http://php.net/manual/en/language.basic-syntax.phpmode.php — CBroe, Jun 03 '17 at 22:57
This with php's error reporting, would have clearly thrown you a parse error. — Funk Forty Niner, Jun 03 '17 at 22:57
@CBroe thank i read through out the article and it really helped — MoRizk, Jun 03 '17 at 23:15
@Fred-ii- thank you fred i'll check more about the error reporting — MoRizk, Jun 03 '17 at 23:15

score 0 · Answer 1 · answered Jun 03 '17 at 22:54

0

You're echoing a string in PHP, and using <?php... inside that string.

Solution:

echo "<a href='secondPage.php?id=" . $i . "'>Link1</a>";

id=$i will also work, because you can include variables directly in double-quoted strings.

answered Jun 03 '17 at 22:54

rybo111

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score 0 · Answer 2 · answered Jun 03 '17 at 22:54

0

You're echoing the PHP code itself as a string. You don't need to put PHP code inside of PHP code. Just concatenate the values you want to echo:

echo '<a href="secondPage.php?id=' . $i . '">Link1</a>';

answered Jun 03 '17 at 22:54

David

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score 0 · Answer 3 · answered Jun 03 '17 at 22:56

because you already started an echo statement so you don't need to add another PHP starting and ending tags. just check my code below and try it.

<?php  
           for ($i=0; $i < 10 ; $i++) { 
             echo "<a href='secondPage.php?id=".$i."'>Link1</a>";
    } ;

?>

score 0 · Accepted Answer · answered Jun 03 '17 at 22:57

<?php and ?> tags indicate to the PHP preprocessor that anything inside them is code and needs to be parsed, everything outside is just text PHP doesn't touch.

Inside the <?php tag, "<?php" string has no special meaning, so is printed. You do not need to open and close tags all the time, try this:

</head>
<body>
    <?php
        $linkName = "Second Page";
        $id = 5;

        for ($i = 0; $i < 10 ; $i++) { 
            echo "<a href='secondPage.php?id=$i;'>Link1</a>";
        };
    ?>
</body>

thank you , it really slipped my eye's but thank you very much — MoRizk, Jun 03 '17 at 23:17

PHP loop is not throwing the correct link

4 Answers4