duanmanmian7589 2018-05-16 11:27
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如何在php中的自动完成文本框中运行两个或更多查询?

code:

<?php
    session_start;
    include('config.php'); 
    $student_id = $_SESSION['student_id'];

    $searchTerm = $_GET['term'];
    $query = "SELECT company_name FROM company WHERE company LIKE '%".$searchTerm."%' ORDER BY company_name ASC";
    $query .= "SELECT key_skills FROM skill_master WHERE key_skills LIKE '%".$searchTerm."%' ORDER BY key_skills ASC";
    $result = mysqli_multi_query($con,$query);
    while($row = mysqli_fetch_assoc($result)) 
    {
        $data[] = $row['company_name'];
        $data[] = $row['key_skills'];
    }
    echo json_encode($data);
?>

I am using autocomplete suggestion box when I use single query in the above code it work perfectly and show me result but now I want to run two query to get different data in a single autocomplete box but when I run this file it show me "null" output. So, How can I fix and get two different table data in single box ?Please help me.

Thank You

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2条回答 默认 最新

  • duanji1026 2018-05-16 11:48
    关注

    The below code was modified to get results from two queries.

    Try this:

    <?php
        session_start();
        include('config.php'); 
        $student_id = $_SESSION['student_id'];
        $data = array();
    
        $searchTerm = $_GET['term'];
        $query = "SELECT company_name FROM company WHERE company LIKE '%".$searchTerm."%' ORDER BY company_name ASC;";
        $query .= "SELECT key_skills FROM skill_master WHERE key_skills LIKE '%".$searchTerm."%' ORDER BY key_skills ASC;";
    
        mysqli_multi_query($con,$query);
        $result = mysqli_store_result($con);
    
        while($row = mysqli_fetch_assoc($result)) 
        {
            $data[] = $row['company_name'];
        }
    
        mysqli_free_result($result);
        mysqli_next_result($con);
    
        $result = mysqli_store_result($con);
    
        while ($row = mysqli_fetch_assoc($result)) {
            $data[] = $row['key_skills'];
        }       
    
        mysqli_free_result($result);
        mysqli_close($con);
        echo json_encode($data);
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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