PHP case switch statement with variable set in drop-down

Asked Apr 19 '17 at 14:47

Active Apr 19 '17 at 15:01

Viewed 199 times

HI can someone tell me where I'm going wrong with this code. For some reason the variable $BlogType; isn't going through the Switch statement. This variable is set using a HTML drop-down and I have tested that its value changes accordingly with the user choice using the echo statement at the beginning of the code.

But It doesn't seem to be going past the first 'break;' in the statement, as the echo statement in the Switch code below isn't printed:

echo "<br /> value in case " . $BlogType;
echo "<br /> Blog " . $BlogType;  // OK, value changes with dropdown choice



function getParameter($param, $defaultValue) {
        if (array_key_exists($param, $_GET)) {
        $value=$_GET[$param];
        return isSet($value)?$value:$defaultValue;
        }
        return $defaultValue;
    }

    $BlogType = getParameter("blog_type", "defaultValue");
    $byDate = getParameter("by_date", "defaultValue");


$loopArr = array();
switch ($BlogType) {

    case 0:
        $loopArr = $all_posts;                                                  
        break;

    case 1: 
        $loopArr = $music_matters->posts;
        echo "<br /> value in case " . $BlogType;
        break;

    case 2:
        $loopArr = $staff_student_news->posts;
        break;

    case 3:
        $loopArr = $kbs_news_events->posts;
        break;

    default: 
        $loopArr = array();
}

?>


<?php foreach ($loopArr as $post): {
    {    // code here

edited Apr 19 '17 at 15:01

asked Apr 19 '17 at 14:47

wilky

http://stackoverflow.com/a/7407500/4916265 – JustOnUnderMillions Apr 19 '17 at 14:48
Try to do `switch (trim($BlogType))`, and you have not show all relevant code to help you more here. – JustOnUnderMillions Apr 19 '17 at 14:51
You need to put brackets around things with more than one thing: `case 1: { $loopArr = $music_matters->posts; echo "
value in case " . $BlogType; } break;` – Evan Carslake Apr 19 '17 at 14:52
2

@EvanCarslake - Brackets are not needed in this case. – Cyclonecode Apr 19 '17 at 14:56
You need to show how you are generating your select element. – Cyclonecode Apr 19 '17 at 14:56
Where are you setting `$BlogType` ? – CD001 Apr 19 '17 at 14:58
have added more code – wilky Apr 19 '17 at 15:00
most likely it is a switch/case nonstrict comparison issue – Deadooshka Apr 19 '17 at 15:00
if you add "var_dump($BlogType);" on the line just before "$loopArr = array();", what does the output say? – trapper_hag Apr 19 '17 at 15:09
hi ti says C:\wamp\www\applicant\views\index.php:82:string 'defaultValue' (length=12) – wilky Apr 19 '17 at 15:13
cast the `$BlogType` to integer – Deadooshka Apr 19 '17 at 15:15
..when the page first loads it says below. could that be the problem. C:\wamp\www\applicant\views\index.php:82:string '' (length=0) – wilky Apr 19 '17 at 15:16
In what way would you cast it? – wilky Apr 19 '17 at 15:18
still no sucess – wilky Apr 19 '17 at 15:23
for instance, the `"defaultValue" == 0` is `true`, but `"defaultValue" == "0"` is `false` – Deadooshka Apr 19 '17 at 15:29

PHP case switch statement with variable set in drop-down

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