how to make ajax receive multiple data from php, and one of the data is in a loop

Question

I know JSON could help me to process multiple data from php by storing all data in an array. But what about one of the data is in a loop? See the code below:

$data1='<p>hehe</p>';
$data2='<h1>wawa</h1>'; 
$backarr=array($data1,$data2);

echo json_encode($backarr);

Now I have something that want to be data3

$sql="SELECT username FROM users WHERE username != '$username'";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array( $result )) {
    $names = $row['username'];
    echo '<p>'.$names.'</p>';
}

Is it possible to do that? I want the ajax to receive all of the three data, how to process the data3?

slash197 · Accepted Answer · 2014-03-14 07:21:05Z

2

It's not that hard really:

$names = array();
while($row = mysql_fetch_array( $result )) 
{
   array_push($names, "<p>" . $row['username'] . "</p>");
}

echo json_encode($names);

answered Mar 14, 2014 at 7:21

slash197

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1 Comment

user2049259 Over a year ago

Thank you, but what about other two data? data1 and data2 are also stored in $names?

LoneWOLFs · Accepted Answer · 2014-03-14 07:34:47Z

0

Try this

$backarray = array();
$backarray['data1'] = '<p>hehe</p>';
$backarray['data2'] = '<h1>wawa</h1>';


$sql="SELECT username FROM users WHERE username != '$username'";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array( $result )) {
    $names[] = $row['username'];
}

//Push $names array into $backarray
$backarray['names'] = $names;

var_dump(json_encode($backarray));

which will output something like under with all names in an array.

'{"data1":"<p>hehe<\/p>","data2":"<h1>wawa<\/h1>","names":["Name1","Name2","Name3","Name4","Name5","Name6"]}'

answered Mar 14, 2014 at 7:34

LoneWOLFs

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4 Comments

user2049259 Over a year ago

I have tried all of the answers, they seem work, but I do not know how to pass them in ajax? Thank you.

LoneWOLFs Over a year ago

@user2049259 Assuming responseObject is your response object received from ajax call, responseObject.data1 will give you data1, responseObject.data2 will give you data2 responseObject.names will give you an array of names which you can iterate by using a loop like responseObject.names[1],responseObject.names[2]...

LoneWOLFs Over a year ago

@user2049259 Are you using jQuery?

user2049259 Over a year ago

No, I am not using JQuery, just raw ajax, your answer works perfectly for me, thank you very much! @LoneWOLFs

ThatMSG · Accepted Answer · 2014-03-14 08:22:33Z

You could combine all in one array and then start to encode.

$dataSet = array();
$data1='<p>hehe</p>';
$data2='<h1>wawa</h1>'; 
$dataSet[] = $backarr = array($data1,$data2);


$sql="SELECT username FROM users WHERE username != '$username'"; 
$result = mysql_query($sql) or die(mysql_error()); 
while($row = mysql_fetch_array( $result )) { 
    $names[] = $row['username'];
} 
$dataSet[] = $names;

echo json_encode($dataSet);

EDIT::

And here the jQuery stuff:

jQuery.post('url/script.php',{fieldName : fieldValue},function(res){
   console.log(res);
   var res = jQuery.parseJSON(res);
   console.log(res);
});

Pure AJAX:

<script type="application/javascript">
function loadJSON()
{
   var data_file = "http://www.example.com/data.json";
   var http_request = new XMLHttpRequest();
   try{
      // Opera 8.0+, Firefox, Chrome, Safari
      http_request = new XMLHttpRequest();
   }catch (e){
      // Internet Explorer Browsers
      try{
         http_request = new ActiveXObject("Msxml2.XMLHTTP");
      }catch (e) {
         try{
            http_request = new ActiveXObject("Microsoft.XMLHTTP");
         }catch (e){
            // Something went wrong
            alert("Your browser broke!");
            return false;
         }
      }
   }
   http_request.onreadystatechange  = function(){
      if (http_request.readyState == 4  )
      {
        // Javascript function JSON.parse to parse JSON data
        var jsonObj = JSON.parse(http_request.responseText);

        // jsonObj variable now contains the data structure and can
        // be accessed as jsonObj.name and jsonObj.country.
        document.getElementById("Name").innerHTML =  jsonObj.name;
        document.getElementById("Country").innerHTML = jsonObj.country;
      }
   }
   http_request.open("GET", data_file, true);
   http_request.send();
}
</script>

Thank you, but how to receive the data in ajax? $('data1').innerHTML=myownrequest2result[0]; $('data2').innerHTML=myownrequest2result[1]; $('data3').innerHTML=myownrequest2result[2]; I hope the [2] should be the data3..

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