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returning an element from JSON

Posted on 2013-05-20

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Last Modified: 2013-06-03

Im trying to get just 1 element from a JSON request:-

switch (json_decode($rowNewInput['json'])['type']) {

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The contents of "$rowNewInput['json']" are:-

{
"type" : "combo",
"attribute" : "SQL",
"values" : "Select distinct `First Name` from `staff`;"
}

And all I want is the 'combo' value.

Although I could decode the JSON into a variable and then pull the variable out of the created array, I was hoping to do it all on one line.

However I get an error to saying:-

Parse error: syntax error, unexpected '['

Any suggestions what Im doing wrong?

Thanks in advance.

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Question by:tonelm54

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2 Comments

LVL 110

Accepted Solution

by:Ray Paseur

Ray Paseur earned 500 total points

ID: 391807472013-05-20

JSON uses object-oriented notation, not array notation. Try the -> instead of the brackets and quotes.

Also, trying to "do it all on one line" is a fools errand. That kind of code makes it harder to debug and understand the data structures. Use as many statements as needed, no less, no more!
http://www.laprbass.com/RAY_temp_tonelm54.php

<?php // RAY_temp_tonelm54.php
error_reporting(E_ALL);

$jso = <<<EOD
{
"type" : "combo",
"attribute" : "SQL",
"values" : "Select distinct `First Name` from `staff`;"
}
EOD;

// DECODE THE STRING INTO AN OBJECT
$obj = json_decode($jso);

// ACTIVATE THIS TO SEE WHAT IS IN THE OBJECT
// var_dump($obj);

// WHAT IS IN THE type PROPERTY
var_dump($obj->type);

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Best regards, ~Ray

LVL 34

Expert Comment

by:Slick812

ID: 391822702013-05-20

greetings tonelm54, , not sure about how you "think" about the methods you use in your code, as you have a PHP ['type'] array access for what you have as a javascript object in the javascript object notation string. I have to work in PHP and javascript at the same time to build pages, and often get mixed up in the code syntax for each,
In javascript you are allowed to access OBJECT properties with array methods like - jsobj['type'] - , but in PHP this is not allowed, , I often forget which Object access to use, and put -> in javascript and put a "." in PHP, which causes fatal errors in either one.
If you are used to javascript, you will need to watch your code when changing to PHP code writing, as many things like ['type'] for object access will get you -
Parse error: syntax error, unexpected '['

and -
switch (json_decode($rowNewInput['json'])->type) {
just might work

but if I forget and use -
switch (json_decode($rowNewInput['json']).type) {
it does Not work.

for your own development error trace, I have found that compound abstraction code like -
switch (json_decode($rowNewInput['json'])->type) {
is harder to debug.

I like this better for debugging -
$newIn = json_decode($rowNewInput['json']);
switch ($newIn->type) {
because you can more clearly see what line and method causes the error

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