PHP，SQL限制通过php变量查询 [英] PHP, SQL limit query by php variable

问题描述

定义变量sqlshowvalue的PHP代码

PHP code defining variable sqlshowvalue

$sqlshowvalue = 5;
if(isset($_POST['showmore'])) {
     $sqlshowvalue += 5;
}

所以我连接到数据库，然后当我使用上面刚刚定义的变量在下面运行此SQL查询时，

So I connect to my database and then when I run this SQL query below using the variable that I just defined above,

$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit '$sqlshowvalue'");

所以我使用mysqli作为连接到数据库的方法，它给了我以下错误:

So I am using mysqli as the method to connect to my DB and it gives me the following error:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in ..

之所以给我这个错误，是因为查询中有错误，它必须执行$ sqlshowvalue，因为如果我仅用数字5替换sqlshowvalue(如下所示)，它就可以正常工作: /p>

The reason it gives me this error is because something in my query is wrong and what it has to do is $sqlshowvalue, because if I replace sqlshowvalue with with just the number 5 (like shown below), it works fine:

$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit 5");

所以我只是想知道如何做才能使limit的值是一个可以更改并更新页面的PHP变量.

So I am just wondering what I can do to make it so that the value for the limit is a PHP variable that I can be changed and the page updated.

推荐答案

您尝试过制作

$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit '$sqlshowvalue'");

到

$result = mysqli_query($conn,"SELECT * FROM comments ORDER BY id DESC limit ".$sqlshowvalue);

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