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如何使用php从我的SQL数据库填充下拉列表

[英]how to populate a drop down list from my SQL database using php

i want to populate a drop down list from my sql database using php, but it shows the below error: 我想使用php从我的sql数据库填充一个下拉列表,但显示以下错误:

"Warning: mysql_fetch_array() expects parameter 1 to be resource, object given in C:\\wamp\\www\\Q&A\\signup.php on line 43" “警告:mysql_fetch_array()期望参数1为资源,第43行的C:\\ wamp \\ www \\ Q&A \\ signup.php中给出的对象”

my code is: 我的代码是:

<?php
$sql = "SELECT category_name FROM category ORDER BY RAND() LIMIT 1";
$result=mysqli_query($conn, $sql) or die ("Query to get data from category failed: ".mysql_error());
    while ($row = mysql_fetch_array($result)) {
    $category_name=$row["category_name"];
    echo "<option>" . $category_name . "</option>";
                }
 ?>

It's because you're using old function mysql_fetch_array it should be mysqli_fetch_array . 这是因为您正在使用旧函数mysql_fetch_array所以它应该是mysqli_fetch_array

Also mysql_error should be mysqli_error which accepts connection as single parameter. 另外mysql_error应该是mysqli_error ,它接受连接作为单个参数。

Update your code like this, 像这样更新您的代码,

<?php
$sql = "SELECT category_name FROM category ORDER BY RAND() LIMIT 1";
$result=mysqli_query($conn, $sql) or die ("Query to get data from category failed: ".mysqli_error(conn));
    while ($row = mysqli_fetch_array($result)) {
    $category_name=$row["category_name"];
    echo "<option>" . $category_name . "</option>";
                }
 ?>

I'm assuming you've a valid connection to the database in $conn variable. 我假设您在$conn变量中具有到数据库的有效连接。

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