Question

我正在尝试使用ajax，php和MySql获取结果。但是，我在服务器端脚本上遇到以下错误。

警告：mysqli_select_db（）期望参数1为mysqli，resource   在第10行的D：\ htdocs \ classes \ xxx中给出

警告：mysql_query（）：提供的参数不是有效的MySQL-Link   资源

服务器端代码如下：

<?php
$q = intval($_GET['q']);

$con= mysqli_connect("localhost","root","abcd123") or die ("could not connect to mysql"); 

mysqli_select_db($con,"payrolldb001") or die ("no database"); 

$sql="SELECT substationid,substationcode FROM wms_substation WHERE assemblylineid = '".$q."'";

$result = mysqli_query($con,$sql);


echo "<select>";
while($row = mysqli_fetch_array($result))
  {
  echo "here";
  echo "<option>". $row['substationcode'] . "</option>";
  }
echo "</select>";

mysqli_close($con);
?>

我无法弄清楚我哪里出错了。请帮忙。

Answer 1

试试这个，

$con = mysqli_connect("localhost","root","abcd123","payrolldb001") or die("Error " . mysqli_error($con));
$sql="SELECT substationid,substationcode FROM wms_substation WHERE assemblylineid = '".$q."'";

$result = mysqli_query($con,$sql);
...

或

$con= mysqli_connect("localhost","root","abcd123") or die ("could not connect to mysql"); mysqli_select_db($con,"payrolldb001") or die ("no database");

阅读mysqli_select_db

Answer 2

    ini_set('display_errors', 1);
    ini_set('display_startup_errors', 1);
    error_reporting(E_ALL);
    //Open a new connection to the MySQL server
    $mysqli = new mysqli('localhost','dbu','password','dbname');

    //Output any connection error
    if ($mysqli->connect_error) {
        die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
    }

    //MySqli Select Query
    $results = $mysqli->query("SELECT * FROM users");

    while($row = $results->fetch_assoc()) {
        echo $row["id"];
    }  


    // Frees the memory associated with a result
    $results->free();

    // close connection 
    $mysqli->close();

无法使用MySQLi_connect连接到MySQL

2 个答案: