无法在Xcode

时间:2016-10-29 22:52:12

标签: php xcode

我尝试使用Xcode,php和mysql将数据发送到在线数据库

创建并测试了mysql数据库和PHP文件。

但是我的IOS代码没有将字符串发送到PHP文件......没有错误,它简单无效!

这里的任何人都知道会出现什么问题吗?

以下是iOS代码:

- (IBAction)send:(id)sender {

NSString *post = [NSString stringWithFormat:@"nome=%@", nome.text];
const char *urlUTF8 = [post UTF8String];
NSString *postBody = [NSString stringWithUTF8String:urlUTF8];

NSData *postData = [postBody dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

NSString *postDataLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
[request setURL:[NSURL URLWithString:[NSString stringWithFormat:@"https://myURLgoes.here/myPHPfile.php"]]];

[request setHTTPMethod:@"POST"];
[request setValue:postDataLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:postData];

这是PHP文件:

<?php
  $con=mysqli_connect("MYSITE.COM","USER","PASSWORD","DATABASE");
if (mysqli_connect_errno($con))
{
 echo '{"query_result":"ERROR"}';
}

$nome = $_GET['nome'];


$result = mysqli_query($con,"INSERT INTO user (nome) 
          VALUES ('$nome')");

if($result == true) {
    echo '{"query_result":"SUCCESS"}';
}
else{
    echo '{"query_result":"FAILURE"}';
}
mysqli_close($con);
?>

0 个答案:

没有答案