Error passing php array with json_encode to js handler

Question

I declare php arrays, i fill them with strings taken from query. I have to pass this arrays to a handler of a element in the page. Handler can't read this arrays.

PHP:

    $names = $paths = $infos= [];
    $i = 0;
    while($row = $result->fetch_row())
    {
            $names[$i]= $row[0];
            $paths[$i]= $row[1];
            $infos[$i]= $row[2];
            $i++;
    } 
    ...
    for($i = 1; $i < ($length/8)+1  ; $i++)//arrotonda per eccesso
    {

        $jsNames = json_encode($names);
        $jsInfo  = json_encode($infos);
        $jsPaths = json_encode($paths);
        //echo $jsNames."<br>";
        //echo $jsInfo."<br>";
        //echo $jsPaths."<br>";
        ?>
        <a href="#" onClick="changePhoto(<?php echo $i.','.$jsNames.','.$jsInfo.','.$jsPaths ?>)" class="w3-bar-item w3-white w3-button"> <?php echo $i ?></a>

JS:

function changePhoto(num,names,paths,infos){
 //nothing
 }

Browser error: Uncaught SyntaxError: Unexpected token }

Opening browser debugger: (function(event){changePhoto(2,[ })

I bet you have a problem with quotes here. Your jsons will have " in it. So try to change onClick="..." to onClick='....' — Jeff
– Jeff, Commented Dec 27, 2017 at 11:46
Should the last } be a ] in (function(event){changePhoto(2,[ }) — Nigel Ren
– Nigel Ren, Commented Dec 27, 2017 at 11:49
but why are you including all the values anyway? You really need that?Or could you just make it echo $i.",'".$names[$i]."','".$infos[$i]."','".$paths[$i] — Jeff
– Jeff, Commented Dec 27, 2017 at 11:49

Jeff · Accepted Answer · 2017-12-27 12:09:59Z

2

You will have a problem with quotes here. Your jsons will have some " in it, and your html-attribute is enclosed in ", too. So change

onClick="changePhoto(<?php echo $i.','.$jsNames.','.$jsInfo.','.$jsPaths ?>)"

to

onClick='changePhoto(<?php echo $i.','.$jsNames.','.$jsInfo.','.$jsPaths ?>)'
// note the single-quotes here!

Sidenote:
You should put these lines $jsNames = json_encode($names); outside the loop, as they won't change anymore. But now you're doing the same work several times.

answered Dec 27, 2017 at 12:09

Jeff

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