php 类 stdClass 的对象无法转换为字符串错误
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Object of class stdClass could not be converted to string error
提问by user2363025
I have the following string:
我有以下字符串：
{"Coords":[{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.277720488429026","Longitude":"-9.012038778269686","Timestamp":"Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27770755361785","Longitude":"-9.011979642121824","Timestamp":"Fri Jul 05 2013 12:02:09 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"},{"Accuracy":"65","Latitude":"53.27769091555766","Longitude":"-9.012051410095722","Timestamp":"Fri Jul 05 2013 12:02:17 GMT+0100 (IST)"}],"Scan":"Whatever"}
Which I want to decode in php. The string is obtained via a sql query. See code below:
我想在 php 中解码。该字符串是通过 sql 查询获得的。见下面的代码：
$TrackDetails_Query= "SELECT * FROM Tracks WHERE TrackID='".$TrackNum."' ORDER BY TrackID DESC";
$TrackDetails_Result= mysql_query($TrackDetails_Query) or die (mysql_error());
if (mysql_num_rows($TrackDetails_Result)==0){
echo 'There are no tracks for the number entered';
}
else{
$traces=$row['Traces'];
$decoded_traces=json_decode($traces);
echo $decoded_traces;
}
}
But I am getting the error:
但我收到错误：
Catchable fatal error: Object of class stdClass could not be converted to string
回答by Paul Dixon
You get the error because you are trying to turn a stdClass object into a string, something it doesn't support.
您收到错误是因为您试图将 stdClass 对象转换为字符串，这是它不支持的。
Instead of echo $decoded_tracestry var_dump($decoded_traces)- that will give a diagnostic view of the object you've decoded (which I presume is what you wanted). You should find it looks like this
而不是echo $decoded_traces尝试var_dump($decoded_traces)- 这将给出您解码的对象的诊断视图（我认为这是您想要的）。你应该会发现它看起来像这样
class stdClass#1 (2) {
public $Coords =>
array(5) {
[0] =>
class stdClass#2 (4) {
public $Accuracy =>
string(2) "65"
public $Latitude =>
string(18) "53.277720488429026"
public $Longitude =>
string(18) "-9.012038778269686"
public $Timestamp =>
string(39) "Fri Jul 05 2013 11:59:34 GMT+0100 (IST)"
}
[1] => (more of the same, edited for brevity)
[2] =>
[3] =>
[4] =>
}
public $Scan =>
string(8) "Whatever"
}
By default, json_encodewill create stdClass objects like the above. If you would prefer an associative array, pass trueas the second parameter to json_decode, e.g. $decoded_traces=json_decode($traces, true);
默认情况下，json_encode会像上面一样创建 stdClass 对象。如果您更喜欢关联数组，true请将其作为第二个参数传递给json_decode，例如$decoded_traces=json_decode($traces, true);
As a further aside, although stdClass can't be turned into a string, your own classes can - a class which implements the __toStringmagic method could be converted to a string!
另外，虽然 stdClass 不能转换为字符串，但您自己的类可以 - 实现__toString魔术方法的类可以转换为字符串！
回答by Nabeel Arshad
use json_encode($traces)this will convert the array into the string. json_decode()is used to convert a string into array or object array
使用json_encode($traces)这会将数组转换为字符串。json_decode()用于将字符串转换为数组或对象数组
回答by dakshbhatt21
Try using this
$decoded_traces=json_decode($traces, true);
尝试使用这个
$decoded_traces=json_decode($traces, true);
instead of this
$decoded_traces=json_decode($traces);
而不是这个
$decoded_traces=json_decode($traces);
It will convert your stdClass to array. Note that $decoded_tracesis array then you can use as you require.
它会将您的 stdClass 转换为数组。请注意，这$decoded_traces是数组，然后您可以根据需要使用。
回答by serena feng
It was an old question, still answer to make it clear for someone who just encounters it like me.
老问题了，还是给我这种刚遇到的人说清楚。
try json_encode($traces);
试试 json_encode($traces);
in my case
就我而言
return var_dump(json_encode($result[0]));
the result is:
结果是：
string(34) "{"id":1,"content":"test database"}"
回答by emmanuel agarry
Instead of
代替
$decoded_traces=json_decode($traces);
echo $decoded_traces;
try
尝试
$decoded_traces=json_decode($traces, true);
print_r $decoded_traces;
回答by deceze
$decoded_tracesis an object. You cannot simply echoan object, because that makes no sense.
$decoded_traces是一个对象。你不能简单地echo成为一个对象，因为那毫无意义。
If you want to debug the object, use var_dump($decoded_traces).
如果要调试对象，请使用var_dump($decoded_traces).
