Unable to parse array string

Question

I've generated an array like below.But unable to parse it in json_encode.Am I missing anything here? please let me know.

PHP code :

<?php

$i=0;
$result = shell_exec('scripts/fetch_script.sh');

foreach(preg_split("/((\r?\n)|(\r\n?))/", $result) as $line)
{
    parse_str($line);

}

$stat = "array(";
    $lines = file('parameter.file');
    foreach ($lines as $line) {
        $v = explode('|', strtoupper($line));
        $str = '$v[1]';
        eval("\$str = \"$str\";");
        eval("\$str = \"$str\";");
        $calc = eval("return round(($str),2);");

        $stat .="'$v[0]' => $calc,";


    }
    $stat .=");";
echo json_encode($stat);
?>

But output shows like Below which is not in json format:

"array('QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,'QUERY_CACHE_USAGE' => 0,'DISK_TMP_TABLES_USAGE' => 1.52,);"

$stat is string it is not array, try with var_dump($stat); and test it — JON
– JON, Commented Nov 28, 2017 at 7:18
Used var_dump in script ,still facing this issue. .. .... var_dump($stat); echo json_encode($stat); below is the output. jpst.it/181m8 — Saikumar A
– Saikumar A, Commented Nov 28, 2017 at 8:32
json_encode($stat, true), try to add true in parameter json_encode — dianeryanto
– dianeryanto, Commented Nov 28, 2017 at 8:48
json_encode works for an array, not string. Change your code to make array instead of string — Aman Rawat
– Aman Rawat, Commented Nov 28, 2017 at 9:05
Why are you using eval() for everything? And why is not "script" not return proper JSON to begin with? — tereško
– tereško, Commented Nov 29, 2017 at 23:54

Hassan · Accepted Answer · 2017-11-28 09:10:29Z

1

The array you're trying to convert to JSON has duplicates, which means that it'll only keep the last occurrence of a duplicated key. For example :

var_dump(['foo' => 'bar', 'foo' => 2]); //displays ["foo"] => int(2)

Since you're already using eval you might as well also use it to evaluate the string into an array :

echo json_encode(eval('return ' . $raw), JSON_PRETTY_PRINT);

Do know however that "eval" is one letter away from "evil". Unless you trust the contents of parameter.file, I'd avoid relying on it. I recommend parsing the file manually instead of relying on eval.

answered Nov 28, 2017 at 9:10

Hassan

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