Question

我正努力使我的搜索功能正常工作。它确实起作用，但是结果不正确。例如，我搜索“数据”一词，并显示“计算机程序员”。它假定显示“数据分析器”。

if (isset($_POST['submit-search'])) {
    $search = mysqli_real_escape_string($conn, $_POST['search']);
   // $sql = 'SELECT * FROM professional, job, location WHERE J.JOB_ID = P.PROFESSIONAL_ID AND P.PROFESSIONAL_ID = L.LOCATION_ID AND PROFESSIONAL_NAME LIKE '%$search%'';
    $sql = "SELECT * FROM professional, job, location WHERE JOB_NAME LIKE '%$search%' AND PROFESSIONAL_ID=JOB_ID AND JOB_ID = LOCATION_ID " ;
    $result = mysqli_query($conn, $sql);
    $queryResult = mysqli_num_rows($result); 
    $row = mysqli_fetch_assoc($result);
    $new_result = $row["JOB_NAME"];

    function str_contains($phrase, $word) {
        return strpos($phrase, $word) !== false;
    }

    if (str_contains($new_result,'computer') || str_contains($new_result, 'programmer') || str_contains($new_result, 'computer programmer')) {
        echo "
        <div class= 'expertise'>
        <h2><i class='fa fa-briefcase' aria-hidden='true'></i><a href='computer_programmer.html'> Computer Programmer</a></h2>
        </div>";
    } else if (str_contains($new_result, 'data') || str_contains($new_result, 'analyst') || str_contains($new_result, 'data analyst')) {
        echo "
        <div class= 'expertise'>
        <h2><i class='fa fa-briefcase' aria-hidden='true'></i><a href='data_analyst.html'> Data Analyst</a></h2>
        </div>";
    } else {
        echo 'There are no results matching your search!';
    }
}

Answer 1

您的if陈述当前不正确。当前，您的表达式将始终求值为true，因为非空字符串被认为是真实的。因此，您的if语句看起来更像：

if ($result == '%computer%' || true || true) {

上面的语句将始终在if内运行代码，因为表达式将始终为true（与X ∨ T ≡ T相同）。

相反，您需要重新检查$ result的相等性：

if ($result == '%computer%' || $result == '%programmer%' || $result == '%computer programmer%') {
  // do "Computer Programer" code...
} else if ($result =='%data%' || $result == '%analyst%' || $result == '%data analyst%') {
 // do "Data Analyst" code...
} else {
  // do "no results code"...
}

但是，要使其正常工作，您需要确保$result是您认为的样子，而不是mysqli_result对象

编辑：要检查字符串是否包含computer或analyst等...您不能像在此处那样使用SQL通配符。相反，您可以使用PHP方法strpos。我建议您构建一个辅助函数，该函数将使您可以在if语句中整齐地执行此操作：

function str_contains($phrase, $word) {
    return strpos(strtolower($phrase), $word) !== false;
}

// then in your if statement:
if (str_contains($result,'computer') || str_contains($result, 'programmer') || str_contains($result, 'computer programmer')) {
  // do "Computer Programer" code...
} else if (str_contains($result, 'data') || str_contains($result, 'analyst') || str_contains($result, 'data analyst')) {
 // do "Data Analyst" code...
} else {
  // do "no results code"...
}

Answer 2

您在代码中使用了$ result错误见

    if (isset($_POST['submit-search'])) {
        $search = mysqli_real_escape_string($conn, $_POST['search']);
       // $sql = 'SELECT * FROM professional, job, location WHERE J.JOB_ID = P.PROFESSIONAL_ID AND P.PROFESSIONAL_ID = L.LOCATION_ID AND PROFESSIONAL_NAME LIKE '%$search%'';
        $sql = "SELECT * FROM professional, job, location WHERE JOB_NAME LIKE '%$search%' AND PROFESSIONAL_ID=JOB_ID AND JOB_ID = LOCATION_ID " ;
        $result = mysqli_query($conn, $sql);
        $queryResult = mysqli_num_rows($result); 
        while($row=mysqli_fetch_assoc($result)){
                if ($row['JOB_NAME'] == '%computer%' || '%programmer%' || '%computer programmer%')
                {
                echo "
                <div class= 'expertise'>
                <h2><i class='fa fa-briefcase' aria-hidden='true'></i><a href='computer_programmer.html'> Computer Programmer</a></h2>
                </div>";
                }
                else if ($row['JOB_NAME'] =='%data%' || '%analyst%' || '%data analyst%'){
                    echo "
                <div class= 'expertise'>
                <h2><i class='fa fa-briefcase' aria-hidden='true'></i><a href='data_analyst.html'> Data Analyst</a></h2>
                </div>";
                }
        }

        } else {
            echo 'There are no results matching your search!';
        }

希望这对您有帮助

为什么我的搜索查询结果不理想？

2 个答案: