Question

以下代码未被执行

 (edited)
   <?php error_reporting(E_ALL); ini_set('display_errors', 1);


     print "Hello world!"; 

$con = mysqli_connect($host,$uname,$pwd,$db) or die(mysqli_error());


     $sql1="SELECT * FROM USERS WHERE username= 'aya'");


   $result = mysqli_query($sql1);
 if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
 }



// json response array
$response = array("error" => FALSE);


    $email = $_POST['username'];
    $password = $_POST['password'];


    $sql="SELECT * FROM USERS WHERE username= 'aya'";
    if(mysqli_query($con,$sql))
    {
     echo json_encode($response);
    }



?>

错误出现在这部分中，因为当我删除它时，链接会给出结果

  $sql="SELECT * FROM USERS WHERE username= 'aya'") or die(mysqli_error())";


     $result = mysqli_query($sql);
     if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
      }

Answer 1

有问题的代码有明显的语法和逻辑错误。
如下所示更正您的代码：

 ...
 // stop script execution with error message if db connection fails
 $con = mysqli_connect($host, $uname, $pwd, $db) or die(mysqli_error());

 $sql1 = "SELECT * FROM USERS WHERE username= 'aya'";

 $result = mysqli_query($sql1);
 if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
 }
...

Answer 2

正如你所说，错误出现在这一部分：

$sql="SELECT * FROM USERS WHERE username= 'aya'") or die(mysqli_error())";

正如您在第一行末尾所看到的，您只需删除双引号"; "。您还需要在(之前添加"SELECT，因此您的代码应为：

$sql= ("SELECT * FROM USERS WHERE username= 'aya'");
$result = mysqli_query($sql) or die(mysqli_error());

Answer 3

更正后的代码：

$con = mysqli_connect($host,$uname,$pwd,$db);
$sql="SELECT * FROM USERS WHERE username='aya'";
$result = mysqli_query($sql) or die(mysqli_error());
if ($result && mysqli_num_rows($result) > 0) {
print "Hea!"; 
}

Answer 4

它的额外双引号＆＃39; ＆＃34; ＆＃39;分号前的第1行结束我删除了它。

  $sql="SELECT * FROM USERS WHERE username= 'aya'") or die(mysqli_error());


  $result = mysqli_query($sql);
     if ($result && mysqli_num_rows($result) > 0) {
     print "Hea!"; 
  }

php代码没有给出正确的结果

4 个答案: