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我發送一些信息給運行查詢的php文件,但我也想同時從該php文件中檢索一些信息。 PHP文件執行得很好,但我無法獲得json_encoded對象。無法使用javascript從php文件發送和接收信息
JavaScript函數發送一個字符串和數字的PHP文件:一個使用字符串和數量,並更新表
<?php
include("config.php");
session_start();
$status = $_POST["Status"];
$num = $_POST["TicketNum"];
$newStatus = " ";
if(strcmp($status, "Open") == 0){
$newStatus = "Closed";
}
elseif(strcmp($status, "Closed") == 0){
$newStatus = "Open";
}
$sql = "UPDATE tickets SET Status = \"$newStatus\" where TicketNum = $num ";
$r = $conn ->query($sql) or trigger_error($conn->error."[$sql]");
$sql = "SELECT * FROM tickets where TicketNum = $num";
$result = $conn->query($sql);
while($row = $result->fetch_assoc()){
$data[] = $row;
}
echo json_encode($data);
?>
我怎樣才能找回json_encoded
function open_close(){
var status = encodeURIComponent(SelectedTicket["Status"]);
var ticketNum = encodeURIComponent(SelectedTicket["TicketNum"]);
var info = "Status="+status+"&TicketNum="+ticketNum;
var http3 = createAjaxRequestObject();
if (http3.readyState == 4) {
if (http3.status == 200){
alert("Ticket Updated!"); //This never gets hit
getUpdatedTicket(JSON.parse(http3.responseText));
}
}
http3.open("POST", "openClose.php", true);
http3.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http3.send(info);
}
PHP文件對象在同一個JavaScript函數中?
來自服務器的響應是'http3.responseText'獲取數據。關於AJAX的任何教程都應該解釋這一點。 – Barmar