问题描述

当 photoId 直接在语句上而不是变量时，以下工作完全没有问题.

The following works with no problem at all, when the photoId is directly on the statement and not a variable.

$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "103"') or die(mysqli_error($con));

但是下面的操作不会出错，可能是什么原因导致无法选择.

but the following just won't work with no error, what might be causing this not to select.

$imageid = '103';
$img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
$img_row = mysqli_fetch_array($img_query);
    echo $img_row['img'];

这是一个while循环.

This is inside a while loop.

while($row = mysqli_fetch_array($somequery)){
 $imageid = $row['photoid'];
    $img_query = mysqli_query($con, 'SELECT * FROM imgs WHERE photoid = "$imageid"') or die(mysqli_error($con));
    $img_row = mysqli_fetch_array($img_query);
        echo $img_row['img'];
}

谢谢.

推荐答案

在php中'和"差别很大，查询语法是双引号左右变量周围的查询和单引号......尽管我建议您在查询中使用参数，而不是直接将变量放入查询中

in php a ' and a " are very different and the query syntax is double quote around the query and single quote around variables.. although I would recommend you look at using parameters on your query instead of just putting a variable directly into the query

$imageid = '103';
$query = $con->prepare("SELECT * FROM imgs WHERE photoid = ?");
$query->bind_param('sssd', $imageid);
$query->execute();

这只是它的细节......如果你想了解更多关于连接的信息......错误处理和其他一切阅读DOCS

this is just the nuts and bolts of it... if you want more information about the connection.. error handling and everything else read the DOCS

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