警告:mysqli_query()期望参数1为mysqli

时间:2017-06-23 09:58:20

标签: php mysqli

<?php

$con = mysqli_connect("localhost","root"," Yes","ecommerce");




//getting the categories

function getCats(){

    global $con;

    $get_cats = "select * from categories";

    $run_cats = mysqli_query($con, $get_cats);

    while($row_cats=mysqli_fetch_array($run_cats)){

        $cat_id = $row_cats['cat_id'];
        $cat_title = $row_cats['cat_title'];

         echo "<li><a href= "#">$cat_title</a></li>";


    ;}

}
?>

错误:

  

警告:mysqli_query()要求参数1为mysqli,第16行的C:\ Xampp \ htdocs \ ecommerce \ functions \ functions.php中给出布尔值

3 个答案:

答案 0 :(得分:0)

检查您的连接查询密码,尝试遵循代码..我的事情空间有问题。

<?php

$con = mysqli_connect("localhost","root","Yes","ecommerce") or die("Not connected.");

//getting the categories

function getCats(){

    global $con;

    $get_cats = "select * from categories";

    $run_cats = mysqli_query($con, $get_cats);

    while($row_cats=mysqli_fetch_array($run_cats)){

        $cat_id = $row_cats['cat_id'];
        $cat_title = $row_cats['cat_title'];

         echo "<li><a href= '#'>$cat_title</a></li>";


    }

}
?>

答案 1 :(得分:0)

我只是用我的数据库连接检查你的代码,但它正在工作!正确检查数据库连接。

die添加到DB连接。

$con = mysqli_connect("localhost","user_name","password","db_name") or die("Not connected.");

答案 2 :(得分:-4)

人们在mysqli中犯的最愚蠢的错误是

  

mysqli_query($con,"SELECT * FROM Persons");

。他们忘了变量:

  

$ CON   或者请与我们分享您的代码。也许我们可以在那之后找到错误解决方案。