Error decoding a javascript json from php

Question

I'm having problems decoding from php a json created with javascript JSON.stringfy. Maybe the problem is that the object has arrays. The JSON and the core are:

$jsonString = "{"tabLabels":["tab1","tab2","tab3","tab4","tab5"],"tabBgs":["21","2","3","0","4"],"tabPublico":[0,0,1,1,0],"fuente":"2","size":"17px"}";

$jsonObj = json_decode($jsonString);

echo $jsonObj->obj;
$tabs = $jsonObj->tabPublico
for ($i=0;$i<strlen($tabs);$i++)
{
   echo $tabs[i];
}

The "echoes" don't show anything.

Thank you for your help. Oscar.

Is that your code? If so, your JSON string is badly escaped. — Florent
– Florent, Commented Nov 21, 2013 at 8:59

Shankar Narayana Damodaran · Accepted Answer · 2013-11-21 09:05:47Z

1

These were your mistakes.

Enclose your $jsonString with single quote
You missed a semicolon ; on the 6th line.
You are using strlen() instead of count().

Modified code.

<?php
$jsonString = '{"tabLabels":["tab1","tab2","tab3","tab4","tab5"],"tabBgs":["21","2","3","0","4"],"tabPublico":[0,0,1,1,0],"fuente":"2","size":"17px"}';

$jsonObj = json_decode($jsonString);
$tabs = $jsonObj->tabPublico;
foreach($tabs as $k=>$v)
{
echo $v;
}

OUTPUT:

edited Nov 21, 2013 at 9:05

answered Nov 21, 2013 at 8:59

Shankar Narayana Damodaran

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2 Comments

Oscardrbcn Over a year ago

Thank you for your answers Shankar and Deepu. I've simplified here my code to focus in the problem and I've had a couple of errors, the jsonString is readed from sql it's not a fixed local var like the example, then I don't have the quote errors and semicolons. But your answers helped me a lot, I've had wrong the loop and I've had a wrong echo that spoiled the outpout. Thanks a lot!

Oscardrbcn Over a year ago

(I've voted up offcourse, I don't know anything about the downvote)

웃웃웃웃웃 · Accepted Answer · 2013-11-21 08:59:16Z

1

Try like this.You json was badly escaped and for an array you must use count() to find the length of array to loop through.

$jsonString = '{"tabLabels":["tab1","tab2","tab3","tab4","tab5"],"tabBgs":["21","2","3","0","4"],"tabPublico":[0,0,1,1,0],"fuente":"2","size":"17px"}';
$jsonObj = json_decode($jsonString);
$tabs = $jsonObj->tabPublico;
for($i=0;$i<count($tabs);$i++)
{
   echo $tabs[$i];
}

answered Nov 21, 2013 at 8:59

웃웃웃웃웃

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Comments

Surabhil Sergy · Accepted Answer · 2013-11-21 09:04:39Z

1

$jsonString = '{"tabLabels":["tab1","tab2","tab3","tab4","tab5"],"tabBgs":["21","2","3","0","4"],"tabPublico":[0,0,1,1,0],"fuente":"2","size":"17px"}'

pass JSON string under single quotes.

answered Nov 21, 2013 at 9:04

Surabhil Sergy

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