‘Insert php, can’t pass its variable to Rawjs

  • world2026

    (@world2026)


    7 years, 5 months ago

    Hello, I’m having problem passing PHP variable to javascript(Rawjs).

    I’ve successfully got data from Mysql and display in HTML .

    But my main goal is to assign the data to js variable. Computed like this

    Rawjs

    [wbcr_php_snippet id="8091"]
<script>
var b="[wbcr_php_snippet id="8101"]";
</script>

    wbcr_php_snippet id=”8091″]

    $query= "select meta_value from wp_gf_entry_meta where meta_key=38";

if ($result = mysqli_query($conn, $query)) {

    while ($row = mysqli_fetch_array($result)) {
        $getii = $row[0];      
    }
  mysqli_free_result($result);
}

    [wbcr_php_snippet id=”8101″]
    echo $getii;

    Have done a lot of research but cant figure out why var b in Rawjs always get 0

    • This topic was modified 7 years, 5 months ago by world2026.
    • This topic was modified 7 years, 5 months ago by world2026.
Viewing 1 replies (of 1 total)
  • Plugin Author webcraftic

    (@webcraftic)

    7 years, 4 months ago

    Hi,

    You can use js code directly inside the snippet.

    See example:

    $query= "select meta_value from wp_gf_entry_meta where meta_key=38";

if ($result = mysqli_query($conn, $query)) {

    while ($row = mysqli_fetch_array($result)) {
        $getii = $row[0];      
    }
  mysqli_free_result($result);
}
?>
<script>
var b="<?=$getii?>";
</script>

    Please check your php code so that the $getii variable has the value you need.

    If you are trying to use the shortcode in double-quotes, this can cause problems. You need to escape double quotes inside the shortcode so that you have no problem with that.

    Best regards, Alex

Viewing 1 replies (of 1 total)

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