Question

我正在尝试将SQL数据库链接到一个应用程序，我正在关注的教程中包含了这个SQL代码。当我替换占位符并运行它时，它给了我上面的错误。这是代码（数据库名称，用户名和表名已被替换，相关部分除外）。真实密码也以感叹号结尾。

<?php

// Create connection
$con=mysqli_connect(“localhost”, ”databasename”, ”password!”, ”tablename”);

// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// This SQL statement selects ALL from the table ‘Recipes’
$sql = "SELECT * FROM Recipes";

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
    // If so, then create a results array and a temporary one
    // to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    {
        // Add each row into our results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    }

    // Finally, encode the array to JSON and output the results
    echo json_encode($resultArray);
}

// Close connections
mysqli_close($result);
mysqli_close($con);
?>

完整错误是：

解析错误：语法错误，意外'！'在第4行的/home/content/93/9076293/html/babyfood/service.php

编辑：以下答案有效，但它打开了一个新错误：

Warning: mysqli_connect() [function.mysqli-connect]: (28000/1045): Access denied for user     'databasename'@'184.168.46.96' (using password: YES) in     /home/content/93/9076293/html/babyfood/service.php on line 4
Failed to connect to MySQL: Access denied for user 'databasename'@'184.168.46.96' (using     password: YES)
Warning: mysqli_query() expects parameter 1 to be mysqli, boolean given in /home/content/93/9076293/html/babyfood/service.php on line 16

Warning: mysqli_close() expects parameter 1 to be mysqli, null given in /home/content/93/9076293/html/babyfood/service.php on line 36

Warning: mysqli_close() expects parameter 1 to be mysqli, boolean given in /home/content/93/9076293/html/babyfood/service.php on line 37

Answer 1

您在连接部分使用智能/引号“ ”而不是标准引号"。

应该如下：

$con=mysqli_connect("localhost", "databasename", "password!", "tablename");

解析错误：语法错误，意外'！'在PHP中

1 个答案: