注意：尝试获取非对象的属性 - 也许是字符串？

Question

I would like to get the value of "Records"

When is try this code:

$json = file_get_contents('URL');
var_dump($json);

This is the result:

string(25289) "{ "ProductsSummary": { "Records": 10, "TotalRecords": 2874, "TotalPages": 288, "CurrentPage": 1 }, "Products": [ { "...

When I try this code

$json = file_get_contents('URL');
$obj = json_decode($json);

echo $obj;
echo $obj->{'ProductsSummary'}->{'Records'};
echo $obj->ProductsSummary->Records;
echo $obj[0]->ProductsSummary->Records;
echo $obj->ProductsSummary[0]->Records;
echo $obj->ProductsSummary[1];

The Output is:

{ "ProductsSummary": { "Records": 10, "TotalRecords": 2879, "TotalPages": 288, "CurrentPage": 1 }, "Products": [ { "Last.... }

Notice: Trying to get property of non-object in  ...
Notice: Trying to get property of non-object in  ...
Notice: Trying to get property of non-object in  ...

Answer 1

Since you do not have valid JSON, I'm pretty sure that json_decode() is just failing. As per the manual:

Returns the value encoded in json in appropriate PHP type. Values true, false and null are returned as TRUE, FALSE and NULL respectively. NULL is returned if the json cannot be decoded or if the encoded data is deeper than the recursion limit.

You can verify that with e.g. the is_null() function:

$obj = json_decode($json);
if( is_null($obj) ){
    // Invalid JSON, don't need to keep on working on it
}else{
    // Read data
}

If everything works fine, $obj will be an object, thus feeding echo with it will never yield anything useful:

echo $obj;

You might want to use var_dump() instead.