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我試圖插入到Web服務器上的數據庫。即時通訊使用MYSQL,Appache和PHP通過webservice連接
的Xcode
*-(IBAction)saveRecord:(id)sender{
UITextField *description_txtfield = (UITextField*)[self.tableView viewWithTag:11];
description_string = description_txtfield.text;
NSLog(description_string);
NSURL *url = [NSURL URLWithString:@"http://192.168.1.140/~admin/qw/"];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];
[request setPostValue:description_string forKey:@"product_name"];
[request setDelegate:self];
[request startAsynchronous];
[self.navigationController popToRootViewControllerAnimated:YES];
}
- (void)requestFinished:(ASIHTTPRequest *)request {
if (request.responseStatusCode == 400) {
NSLog(@"Invalid code");
} else if (request.responseStatusCode == 403) {
NSLog(@"Code already used");
} else if (request.responseStatusCode == 200) {
NSLog(@"OK");
} else {
NSLog(@"Unexpected error");
}
}*
- (void)requestFailed:(ASIHTTPRequest *)request
{
NSError *error = [request error];
NSLog(error.localizedDescription);
}
index.php has the following code inside to handle insert.
Function redeem() {
// Check for required parameters
if (isset($_POST["product_name"])) {
// Put parameters into local variables
$rw_app_id = $_POST["product_name"];
// Add tracking of redemption
$stmt = $this->db->prepare("INSERT INTO inventory (product_name) VALUES (?)");
$stmt->bind_param($rw_app_id);
$stmt->execute();
$stmt->close();
}
return false;
}
誰能告訴我做什麼我錯在這裏。我一直收到無效的代碼,這是錯誤的請求400。幫幫我!這讓我瘋狂!
我把它拿出來,但它還沒有經過。我無花果。這是給我一些問題,但我告訴了它,我從這個方法得到一個意想不到的錯誤 - (void)requestFinished:(ASIHTTPRequest *)請求。即時通訊使用IP地址在我的網址導致問題的事實? – user984373 2012-01-08 12:08:25
我怎樣才能確保index.php接收一個值?我可以查詢數據庫以列出所有產品並將其顯示在我的頁面上。那工作找到。似乎index.php頁面沒有在我點擊應用程序上的保存按鈕時收到值。任何方式來檢查這一點? – user984373 2012-01-08 12:14:59
不,使用數字IP地址不是問題。也許你需要從PHP端發回'200'響應,所以你的客戶端的'requestFinished'方法會報告所有的都是成功的? – 2012-01-08 12:17:25