Question

我知道Rand () 的顺序并不是从数据库中抽取随机值的最快方法, 但我的数据库是很小的, 在这一点上; 我只是想让它起作用! 哈哈。这是我的代码 :

include( includes/dbc.php );
$top_query = "SELECT * FROM top WHERE  occasion_id =" . $occasion . "  AND  temperature_id = " . $temperature . "  AND  user_id = " . $user_id . " ORDER BY RAND() LIMIT 1";
$top_result = mysqli_query($dbc, $top_query) or die (  The top SELECT query is broken. );
mysqli_close ($dbc);

while($row= mysqli_fetch_array($top_result)) {
    echo  This top has an id of:  . $row[ top_id ]  .  <br>  ;
    echo  Does this top require pants?  . $row[ needs_pants ]  .  <br> ;
    echo  What s the colour id of this top?  . $row[ colour_id ] .  <br> ;
    echo  $row[ value ];
}

出于某种原因,这不起作用, 当我尝试运行我的阵列时, 我就会白白露面。在我按“ 兰德( ) 限制 1 ” 键按顺序排列之前, 它就起作用了, 但显然我得到了每一个价值, 而不是一个随机值。

有人看到我哪里出错了吗?

Answer 1

您的查询格式错误。它不是 dying , 它只是没有返回结果。

请注意错误的单引号:

$top_query = "SELECT * FROM top WHERE occasion_id =" . $occasion . " AND temperature_id = " . $temperature . " AND user_id = " . $user_id . " ORDER BY RAND() LIMIT 1";

您可以通过输出查询来证明这一点, 然后直接在 MySQL 中运行( 例如通过 PHPMYAdmin ) :

echo $top_query;

功能如 mysqli_error () 和 mysqli_num_rows () 帮助确定查询结果。

为了在路上移动你, SQL应该更像:

$top_query = "SELECT * FROM top WHERE occasion_id = " . $occasion . " AND temperature_id = " . $temperature . " AND user_id = " . $user_id . " ORDER BY RAND() LIMIT 1";

一些补充说明:

Read up on SQL Injection.

Familiarize yourself with when you need to quote literal values in MySQL. In your case these are all ID columns (theoretically integers) and therefore don t need to be quoted.

Be mindful not to prematurely close your MySQL connection. PHP will actually do this automatically at the end of the script.

As commented by webbiedave, also look into Prepared Statements.

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