How to parse json in php without converting into array?

Question

{
    "\u0000*\u0000collection_key": "rows",
    "\u0000*\u0000internal_gapi_mappings": [],
    "cacheHit": true,
    "\u0000*\u0000errorsType": "Google_Service_Bigquery_ErrorProto",
    "\u0000*\u0000errorsDataType": "array",
    "jobComplete": true,
    "\u0000*\u0000jobReferenceType": "Google_Service_Bigquery_JobReference",
    "\u0000*\u0000jobReferenceDataType": "",
    "kind": "bigquery#queryResponse",
    "pageToken": null,
    "\u0000*\u0000rowsType": "Google_Service_Bigquery_TableRow",
    "\u0000*\u0000rowsDataType": "array",
    "\u0000*\u0000schemaType": "Google_Service_Bigquery_TableSchema",
    "\u0000*\u0000schemaDataType": "",
    "totalBytesProcessed": "0",
    "totalRows": "1",
    "\u0000*\u0000modelData": {
        "schema": {
            "fields": [{
                "name": "total",
                "type": "INTEGER",
                "mode": "NULLABLE"
            }]
        },
        "jobReference": {
            "projectId": "qa-bigquery",
            "jobId": "job_y_qXq9mbzrnHzQZf_CUCr8itgmA"
        },
        "rows": [{
            "f": [{
                "v": "666750"
            }]
        }]
    },
    "\u0000*\u0000processed": []
}

I have above output from my script I need to traverse through rows & get value {"v":"666750"}] ? Can you please help ?

You don't want to decode it? That's the best and easiest way... — Antony Thompson
– Antony Thompson, Commented Apr 28, 2017 at 9:26
It might be better to look at how you generate this JSON String or how you call the API that returns it to you. — RiggsFolly
– RiggsFolly, Commented Apr 28, 2017 at 9:49

RiggsFolly · Accepted Answer · 2017-04-28 10:01:09Z

This is not particularly elegant, but it does seem to work.

First I removed all the \u0000*\u0000 from the string and then just did a normal json_decode()

$s = '{
    "\u0000*\u0000collection_key": "rows",
    "\u0000*\u0000internal_gapi_mappings": [],
    "cacheHit": true,
    "\u0000*\u0000errorsType": "Google_Service_Bigquery_ErrorProto",
    "\u0000*\u0000errorsDataType": "array",
    "jobComplete": true,
    "\u0000*\u0000jobReferenceType": "Google_Service_Bigquery_JobReference",
    "\u0000*\u0000jobReferenceDataType": "",
    "kind": "bigquery#queryResponse",
    "pageToken": null,
    "\u0000*\u0000rowsType": "Google_Service_Bigquery_TableRow",
    "\u0000*\u0000rowsDataType": "array",
    "\u0000*\u0000schemaType": "Google_Service_Bigquery_TableSchema",
    "\u0000*\u0000schemaDataType": "",
    "totalBytesProcessed": "0",
    "totalRows": "1",
    "\u0000*\u0000modelData": {
        "schema": {
            "fields": [{
                "name": "total",
                "type": "INTEGER",
                "mode": "NULLABLE"
            }]
        },
        "jobReference": {
            "projectId": "qa-bigquery",
            "jobId": "job_y_qXq9mbzrnHzQZf_CUCr8itgmA"
        },
        "rows": [{
            "f": [{
                "v": "666750"
            }]
        }]
    },
    "\u0000*\u0000processed": []
}';

$s1 = str_replace(array('\u0000*\u0000'), '', $s);
$j = json_decode($s1);
//echo json_last_error_msg();
//print_r($j);

// you now have a PHP object as defined by the original string

echo $j->modelData->rows[0]->f[0]->v;

RESULT

Tomasz · Accepted Answer · 2017-04-28 09:26:22Z

0

Take a look on json_decode() function.

<?php

$jsonString = '{"*collection_key":"rows","*internal_gapi_mappings":[],"cacheHit":true,"*errorsType":"Google_Service_Bigquery_ErrorProto","*errorsDataType":"array","jobComplete":true,"*jobReferenceType":"Google_Service_Bigquery_JobReference","*jobReferenceDataType":"","kind":"bigquery#queryResponse","pageToken":null,"*rowsType":"Google_Service_Bigquery_TableRow","*rowsDataType":"array","*schemaType":"Google_Service_Bigquery_TableSchema","*schemaDataType":"","totalBytesProcessed":"0","totalRows":"1","*modelData":{"schema":{"fields":[{"name":"total","type":"INTEGER","mode":"NULLABLE"}]},"jobReference":{"projectId":"qa-bigquery","jobId":"job_y_qXq9mbzrnHzQZf_CUCr8itgmA"},"rows":[{"f":[{"v":"666750"}]}]},"*processed":[]}';
$jsonObject = json_decode($jsonString);

var_dump($jsonObject->modelData->rows->f->v)

answered Apr 28, 2017 at 9:26

Tomasz

5,2222 gold badges37 silver badges42 bronze badges

2 Comments

RiggsFolly Over a year ago

His problem is (if you test his actual json string) that it will not decode and you get an error The decoded property name is invalid

Mayank pandey Over a year ago

Yes @RiggsFolly I am getting exception as "Fatal error: Cannot access property started with '\0' in /Applications/XAMPP/xamppfiles/htdocs/BQ.php on line 48"

Collectives™ on Stack Overflow

How to parse json in php without converting into array?

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related