mysqli num行无法工作的php问题

时间:2018-09-02 19:23:28

标签: php ajax

我对此功能mysqli_num_rows()有问题,因为有时对我来说效果很好,有时它让我头疼,我真的不知道这是怎么回事

这是我的PHP文件:

<?php  

    include 'dbh.php';

    function get_font_class() {
        global $conn;
        if ( isset($_POST['dataSearchBox']) ) {
            $searchKeyword = $_POST['dataSearchBox'];
            $sql = ' SELECT * FROM fontawesomeicons WHERE name LIKE ' .$searchKeyword. '%; ';
            $result = mysqli_query($conn, $sql);
            if ( mysqli_num_rows( $result ) > 0 ) {
                while ( $row = mysqli_fetch_assoc( $result ) ) {
                    echo '<i class=' .$row['class']. '></i>';
                }
            } else {
                echo "There Are No Icons At The Moment!";
            }
        }   
    }
    get_font_class();

2 个答案:

答案 0 :(得分:0)

在课程属性之前和之后使用双引号  

    include 'dbh.php';

    function get_font_class() {
        global $conn;
        if ( isset($_POST['dataSearchBox']) ) {
            $searchKeyword = $_POST['dataSearchBox'];
            $sql = ' SELECT * FROM fontawesomeicons WHERE name LIKE ' .$searchKeyword. '%; ';
            $result = mysqli_query($conn, $sql);
            if ( mysqli_num_rows( $result ) > 0 ) {
                while ( $row = mysqli_fetch_assoc( $result ) ) {
                    echo '<i class="' .$row['class']. '"></i>';
                }
            } else {
                echo "There Are No Icons At The Moment!";
            }
        }   
    }
    get_font_class();

答案 1 :(得分:-2)

将sql写为

$sql = "SELECT * FROM fontawesomeicons WHERE name LIKE '%".$searchKeyword."%'" ;
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