如何使用getJson将变量发送到php文件

Question

Here is the javascript code that has a "no" in the datastring and i want to pass that datastring to data.php using getJson So that in the end i can get json as a response. My data.php code is working but javascript code is not working so please help me.

Javascript code

dataString = 'no='+ no; 

$.getJSON("data.php", dataString,function(response){

        console.log(response);
  });

php code

$no=$_POST['no'];

$headers = array(
'Content-Type: application/json',
);

$url = 'https://www.outpan.com/api/get-product.php?barcode='.$no.'&apikey=318c91a99b132a29dda536d63e7fb8b4';


$ch = curl_init();


curl_setopt($ch, CURLOPT_URL, $url);

curl_setopt($ch, CURLOPT_POST, false);

curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);


curl_setopt($ch, CURLOPT_RETURNTRANSFER, true );


curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);




 $result = curl_exec($ch);



curl_close($ch);

//$result_arr = json_decode($result, true);

echo result;

Answer 1

One possibility is response is not valid JSON (I am not sure, since you seemed to have tried printing the json_decoded version ).
Try printing out the error by using error which is called when no valid JSON is received -

$.getJSON("data.php", dataString,function(response){
    console.log(response)
}).error(function(jqXHR, textStatus, errorThrown){ 
    console.log(textStatus+" - "+errorThrown); 
}); 

If an error is logged, then checkout the curl response in php and make sure it is valid json.

EDIT
This edit is assuming that he php response is invalid JSON as observed from the comments.
Looking at your php code, you have used this -

$headers = array(
'Content-Type: application/json',
);

Try this instead -

header('Content-Type: application/json');

EDIT 2
Looking at your php code again, there seems to be an error while printing -

echo result;

It should be -

echo $result;

Answer 2

使用$ _GET在您的php代码中，您现在正在使用$ _POST

Answer 3

You should $_GET in your php code or change code like this :

$.getJSON("data.php", { no: no },function(response){

    console.log(response);
  });

In php use,

$no=$_REQUEST['no'];