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在php中解码JSON文件时出现无效参数错误

json php

0

Martin Mbae · 技术社区 · 7 年前

<?php
  $content =  file_get_contents('result.json');

  $contendecoded = json_decode($content);

  print_r($contendecoded);

  foreach ($contendecoded as $each) {
    # code...
    echo '<li>' .$each->Receiver. '</li>';
  }
  ?>

这是result.json文件

{"Sender":"254705537065","Receiver":"7528452889","Amount":"1","FName":"Martinique","AccessToken":"8XRTHN59NCHvUGAASGbK6IcCzYcn"},
{"Sender":"254705537065","Receiver":"6584238686","Amount":"2","FName":"Phillipines2","AccessToken":"O4wBFWPmFA8ayKGYahhpdCAW97mg"},
{"Sender":"254705537065","Receiver":"6584238686","Amount":"2","FName":"Phillipines2","AccessToken":"O4wBFWPmFA8ayKGYahhpdCAW97mg"},
{"Sender":"254705537065","Receiver":"6584238686","Amount":"36","FName":"Phillipines3","AccessToken":"O4wBFWPmFA8ayKGYahhpdCAW97mg"},
{"Sender":"254705537065","Receiver":"6584238686","Amount":"36","FName":"Phillipines3","AccessToken":"O4wBFWPmFA8ayKGYahhpdCAW97mg"}

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\website\receive.php on line 8

错误所指的第8行是

foreach ($contendecoded as $each) {

我该怎么解决这个问题?

3 回复 | 直到 7 年前

1

shawn 7 年前

文件内容不是有效的JSON。

你可以参考这个问题(我也回答了) Decoding multiple JSON objects in PHP

function json_decode_multi($s, $assoc = false, $depth = 512, $options = 0) {
    if(substr($s, -1) == ',')
        $s = substr($s, 0, -1);
    return json_decode("[$s]", $assoc, $depth, $options);
}

$content =  file_get_contents('result.json');
$contendecoded = json_decode_multi($content);
print_r($contendecoded);

2

1

Murzid 7 年前

请再次发布您的PHP代码

3

1

Suraj Kumar zip 7 年前

我认为在JSON文件中需要一个括号对,如下所示

[
{},
{}
]