Question

您好我试图在php中编写搜索功能，但我总是得到0结果。

<?php
$mysqli = new mysqli('localhost', 'root', 'testing123', 'test') 
if ($mysqli->connect_error) {
    die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
} 
//collect
$output = '';
if(isset($_POST['search'])){
    $searchq = $_POST['search'];
    $searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
    $query = mysqli_query($mysqli, "SELECT * FROM Users WHERE UsernameLIKE '%$searchq%'");
    $count = mysqli_num_rows($query);
    // if the $result contains at least one row
    if ($query->num_rows > 0) {
        // output data of each row from $result
        while($row = $query->fetch_assoc()) {
        }
    }
    else {
        echo '0 results';
    }
}    
?>

任何想法如何解决？

编辑1.即使这被注释，这仍然不允许它被解决。

if(!empty($_POST['search'])){
    $searchq = $_POST['search'];
    //$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
    $query = mysqli_query($mysqli, "SELECT * FROM Users WHERE Username LIKE '$searchq%'");
    $count = mysqli_num_rows($query);
    // if the $result contains at least one row
    if ($query->num_rows > 0) {
        // output data of each row from $result
        while($row = $query->fetch_assoc()) {
        }
    }
    else {
        echo '0 results';
    }
}

Answer 1

为了匹配学校，学校名称或学校名称1

您的查询需要读取并删除起始%并保留尾随%

WHERE SchoolName LIKE '$searchq%'"

现在，它找不到学校名称和学校名称1的原因是因为它包含空格而你的正则表达式正在替换那些空格。

$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

注释掉该行或修改正则表达式。

通过评论，您可以恢复使用：

WHERE SchoolName LIKE '%$searchq%'"

PHP MYSQL搜索总是带回0结果

1 个答案: