如何在另一个查询的while循环内运行sql查询

问题描述：

当photoId直接在语句上而不是变量时，以下内容完全没有问题.

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));

但是以下内容将不会有任何错误，这可能是导致未选择该内容的原因.

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'];

这是在while循环内.

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.

答

在php中，'和"有很大的不同，查询语法在查询周围是双引号，在变量周围是单引号.建议您考虑在查询中使用参数，而不是直接将变量直接放入查询中

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();

这只是它的基本内容...如果您想了解有关连接的更多信息..错误处理和其他所有内容，请阅读

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

如何在另一个查询的while循环内运行sql查询

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