你可以在一個陣列結合所有,然後開始編碼。
$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);
編輯::
這裏jQuery的東西:
jQuery.post('url/script.php',{fieldName : fieldValue},function(res){
console.log(res);
var res = jQuery.parseJSON(res);
console.log(res);
});
純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>
我已經嘗試了所有的答案,他們似乎工作,但我不知道如何將它們傳遞給ajax?謝謝。 – user2049259
@ user2049259假設'responseObject'是你從ajax調用收到的響應對象,'responseObject.data1'會給你'data1','responseObject'.data2會給你'data2'' responseObject.names'會給你一個數組名稱,你可以通過循環使用像'responseObject.names [1]','responseObject.names [2]''... – LoneWOLFs
@ user2049259您是否使用jQuery? – LoneWOLFs