Question

I have a function. in that function in have a select query. another query should return the number of results that will be in the first query. I have this code:

    function name ($f){  
     global $db;
     ...
     $results = "SELECT `a` FROM `b` WHERE $where"; 
     $results_num = ($query = mysqli_query($db, $results)) ? mysqli_num_rows($results) : 0; 

     echo $results;
     echo $results_num;
    }

将重复:

SELECT `a` FROM `b` WHERE `keywords` LIKE  %abc% 0

什么只是$results,但不是$results_num/code>? 我不理解为什么要显示<代码>echo$results_num? 为什么在底有<0><0>><0>>> > ,这样,如果有人能够给我建议解决我真的希望的话。非常感谢。

Answer 1

首先,$results_num为零,正在产出中(这就是为什么在产出结束时有<0>)。这是因为该法典是错误的:

$results_num = ($query = mysqli_query($db, $results)) ? mysqli_num_rows($results) : 0;

略微细分:

$query = mysqli_query($db, $results);
$results_num = $query ? mysqli_num_rows($results) : 0;

第二行实际上应当是:

$results_num = $query ? mysqli_num_rows($query) : 0;

您需要将电离层通过到mysqli_num_rows(,而不是通过K.。

Answer 2

$results_num is 0. $results is - SELECT a FROM b WHERE keywords LIKE %abc% try this

echo "Results is $results <br />";
echo "Results_num is $results_num <br />";

并且一切都表明:

Answer 3

。您可以转至mysqli_num_rows(<>/code>,就像你再做一样,期望通过mysqli_query(返回的一套结果识别码。阁下:<代码> (你可能会改变你的变量名称,使之更有意义,即图瓦卢妇女委员会)。

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