Question

我正在尝试遍历我的数据库并检查用户是否已经存在于另一个表中。如果他们这样做，那么我想增加一个值，如果他们不想，那么我想添加用户。

当我运行下面的代码时，它会愉快地遍历所有结果：

＆＃13;

<?php

$servername = "p:10*********";
$username = "*******";
$password = "*******";
$dbname = "******";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

echo $result->num_rows;
 while($row = $result->fetch_assoc()) {




  
$user= $row['user'];
$time= $row['time'];
$id= $row['id'];

echo $id;

echo $user;



    
}

} else {
    echo "0 results";
}
$conn->close();



?>

＆＃13;

但是，当我添加SQL以检查它们是否存在于另一个表中时，循环不再正常运行并且每次都回显相同的用户。

＆＃13;

<?php

$servername = "*******";
$username = "******";
$password = "********";
$dbname = "*****";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM payroll WHERE user != ' ' ";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

echo $result->num_rows;
 while($row = $result->fetch_assoc()) {




  
$user= $row['user'];
$time= $row['time'];
$id= $row['id'];

echo $id;

echo $user;
$sql = "SELECT * FROM smsreport WHERE user = '$user'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
  echo "found";
    
} else {
   echo "USER NOT FOUND";
}




    
}

} else {
    echo "0 results";
}
$conn->close();



?>

＆＃13;

任何帮助都非常感激。

Answer 1

睁眼：重命名内部$result变量。它正在翻阅第一个$result。

可能是问题所在。虽然没有测试过。

PHP循环结果

1 个答案: