我試圖開發使用ajax的職位和列表選擇更新後者從MySQL數據庫的響應的應用程序,但列表顯示爲空,可以將部分一個扶從這個我出去,請.....爲的.js安裝webOS的列表選擇的選擇與AJAX JSON響應動態
代碼:爲PHP
SecondAssistant.prototype.setup = function() {
this.selectorChanged = this.selectorChanged.bindEventListener(this);
Mojo.Event.listen(this.controller.get('firstselector'), Mojo.Event.propertyChange, this.selectorChanged);
this.names = [];
try {
new Ajax.Request('http://localhost/projects/testingasdf.php', {
method: 'post',
parameters: {
'recs': getallrecords,
'q': q
},
evalJSON: 'true',
onSuccess: function(response){
var json = response.responseJSON;
var count = json.count - 1;
for(i=0; i<count; i++){
this.names.push({
label: json[i].name,
value: '0'
});
}
this.controller.modelChanged(this.model);
}.bind(this),
onFailure: function(){
Mojo.Controller.errorDialog('Failed to get ajax response');
}
});
}
catch (e){
Mojo.Controller.errorDialog(e);
}
this.controller.setupWidget("firstselector",
this.attributes = {
label: $L('Name'),
modelProperty: 'currentName'
},
this.model = {
choices: this.names
}
);
};
代碼:
<?php
header('Content-type: application/json'); // this is the magic that sets responseJSON
$conn = mysql_connect('localhost', 'root', '')// creating a connection
mysql_select_db("test", $conn) or die('could not select the database');//selecting database from connected database connection
switch($_POST['recs'])
{
case'getallRecords':{
$q = $_POST['q'];
//performing sql operations
$query = sprintf("SELECT * FROM user WHERE name= $q");
$result = mysql_query($query) or die('Query failed:' .mysql_error());
$all_recs = array();
while ($line = mysql_fetch_array($result,MYSQL_ASSOC)) {
$all_recs[] = $line;
}
break;
}
}
echo json_encode($all_recs);
// Free resultset
mysql_free_result($result);
// closing connection
mysql_close($conn);
?>
我試圖相同,但B =使用遠程託管服務器..本地主機是不是爲我工作:( – 2011-06-21 12:48:29