[英]json_decode not working in php
請檢查我下面的代碼,並讓我知道我做錯了什么,我只想解碼json並在瀏覽器上以數組形式打印
<?php
$json = '{
"Token" : "xxx-xxx-xxx",
"ID": "1",
"Recipients": [
{
"Recipient_ID": "XX",
"From_Name": "XXX",
"From_Email": "XXX",
"To_Name": "XXX",
"To_Email": "XXX",
"Subject": "XXX",
"Message": "XXX",
"Attachments": [
{
"File_Name": "XXX",
"File_Path": "XXX",
}
],
}
],
}';
$input = $json;
print_r(json_decode(stripslashes($input)));
?>
我嘗試使用此json字符串進行在線解碼( http://jsonviewer.stack.hu/ ),並且它可以正常工作,因此json字符串沒有問題。 任何幫助將不勝感激。
<?php
$json = '{
"Token": "xxx-xxx-xxx",
"ID": "1",
"Recipients": [{
"Recipient_ID": "XX",
"From_Name": "XXX",
"From_Email": "XXX",
"To_Name": "XXX",
"To_Email": "XXX",
"Subject": "XXX",
"Message": "XXX",
"Attachments": [{
"File_Name": "XXX",
"File_Path": "XXX"
}]
}]
}';
$input = $json;
var_dump(json_decode(stripslashes($input)));
?>
問題出在您擁有的json格式上。您有一些需要刪除的',',這就是json_decode()無法完成工作的原因。 該函數引發錯誤異常,但是您應該做一些技巧來查看錯誤。 您可以使用此代碼查看錯誤。
switch (json_last_error()) {
case JSON_ERROR_NONE:
echo ' - No errors';
break;
case JSON_ERROR_DEPTH:
echo ' - Maximum stack depth exceeded';
break;
case JSON_ERROR_STATE_MISMATCH:
echo ' - Underflow or the modes mismatch';
break;
case JSON_ERROR_CTRL_CHAR:
echo ' - Unexpected control character found';
break;
case JSON_ERROR_SYNTAX:
echo ' - Syntax error, malformed JSON';
break;
case JSON_ERROR_UTF8:
echo ' - Malformed UTF-8 characters, possibly incorrectly encoded';
break;
default:
echo ' - Unknown error';
break;
}
而且您的json應該是這樣的。
$json = '{
"Token" : "xxx-xxx-xxx",
"ID": "1",
"Recipients": [
{
"Recipient_ID": "XX",
"From_Name": "XXX",
"From_Email": "XXX",
"To_Name": "XXX",
"To_Email": "XXX",
"Subject": "XXX",
"Message": "XXX",
"Attachments": [
{
"File_Name": "XXX",
"File_Path": "XXX"
}
]
}
]
}';
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