Question

我不知道怎么了，我得到一个错误。帮我。预先谢谢你。

PHP代码

<?php
//DATABASE
$con=mysqli_connect("localhost", "root", "pw1234", "mr_system_db");

if(mysqli_connect_errno($con)) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();   
}



//$Today = $_POST['today'];


$now_date = date("Y-m-d");
echo $now_date;
//쿼리
$query = "SELECT MR_Dept_TB_Dept_NO, MR_MtRoom_TB_MtRoom_NO, MR_Info_Stime, MR_Info_Etime, mr_info_date
         FROM mr_info_tb where mr_info_date =  '$now_date'";

错误点！ $结果= mysqli_query（$查询）;

mysqli_query（）至少需要2个参数，给定1个

$result = mysqli_query($query); 

if(!$result )

错误点* mysqli_error（）恰好需要1个参数，给定0个参数

die ( "Database access failed  : "  . mysqli_error());


    while ($row = mysqli_fetch_object($result)) {
        $arr[] = $row;  
    }
    echo json_encode($arr, JSON_UNESCAPED_UNICODE);


    mysqli_free_result($result);
    mysqli_close($con);

？>

Answer 1

您应该更改

$result = mysqli_query($query);至$result = mysqli_query($con,$query);

和

mysqli_error()至mysqli_error($con)

PHP警告：mysqli_query（）至少需要2个参数，即1

1 个答案: