What is the JSON requirement for ios-php to the app?

I get the data out of the database and send it to the web, so the app gets the information:

$result = mysql_query("select * from his");while ($arr = mysql_fetch_assoc($result)) {    $json = json_encode($arr,JSON_UNESCAPED_UNICODE);    echo $json;}

The JSON output to the interface is:
{"hps_userid":"1","hps_userid2":"1"}
Is it my output format problem or the app side problem that the app can't get data now?

Now the app's colleague says it needs this format to

{"No.":[{"hps_userid":"1","hps_userid2":"1"}]}

When searching for multiple data, how do I output it?

Reply content:

I get the data out of the database and send it to the web, so the app gets the information:

$result = mysql_query("select * from his");while ($arr = mysql_fetch_assoc($result)) {    $json = json_encode($arr,JSON_UNESCAPED_UNICODE);    echo $json;}

The JSON output to the interface is:
{"hps_userid":"1","hps_userid2":"1"}
Is it my output format problem or the app side problem that the app can't get data now?

Now the app's colleague says it needs this format to

{"No.":[{"hps_userid":"1","hps_userid2":"1"}]}

When searching for multiple data, how do I output it?

"Now the app side colleagues say that need this format can", "just can", haha ...
I've heard a similar story before when I worked with someone else, but then I learned about Android development ...
Mom eggs, only to find, obviously they parse JSON code not how can write, also blame I give the format is wrong ...

As long as the output is the standard JSON format, the information is complete, the organization is correct, there is no resolution of the problem.

{"No.":[{"hps_userid":"1","hps_userid2":"1"}]}

The existence of this "no" is not necessary (just because they are not very good at JSON parsing).
In other words, "when searching for multiple data, how do you output it?" "
This kind of problem should be considered right from the start, okay?

[    {        "hps_userid": "1",        "hps_userid2": "1"    },    {        "hps_userid": "2",        "hps_userid2": "3"    },    {        "hps_userid": "4",        "hps_userid2": "5"    }]

Take more than one piece of data, save it in an array, and then turn the array into JSON, that's it.
How can you say your code is like this ...

$result = mysql_query("select * from his");while ($arr = mysql_fetch_assoc($result)) {    $json = json_encode($arr,JSON_UNESCAPED_UNICODE);    echo $json;}

Do you know how many times the while loop? Each time the loop echoes, the final output is something like this:

{"hps_userid":"1","hps_userid2":"1"}{"hps_userid":"1","hps_userid2":"1"}{"hps_userid":"1","hps_userid2":"1"}{"hps_userid":"1","hps_userid2":"1"}{"hps_userid":"1","hps_userid2":"1"}{"hps_userid":"1","hps_userid2":"1"}

This kind of connection is not a JSON string at all ...
。。。。。。
As follows:

$result = mysql_query("select * from his");$Temp = array();$row = mysql_fetch_assoc($result);while ($row) {    $Temp[] = array("hps_userid"=>$row["hps_userid"],"hps_userid2"=>$row["hps_userid2"]);    $row = mysql_fetch_assoc($result);}die(json_encode($Temp));//或者你就听你们那个逗比APP工程师的：//die(json_encode(array("No."=>$Temp)));

Here is the need to loop, so you're going to tease the JSON for you than the engineer, which is needed in [{xxx:1},{xxx:2}] this format

