搜索表单多输入

Question

我有一个搜索表单有2个选择字段和一个输入，总共有3个选项，所以我创建了一些if语句，具体取决于每个字段的设置，它有自己的查询，但它运行不好，是所有的错误，结果不正确，它与语句查询混淆，它不是正确的。

这里是我的搜索表单结果代码：）

$keywords = $_GET["Keywords"]; 
      $location = $_GET['Location']; 
      $jobtype = $_GET["Category"]; 


       if (isset($location) && empty($jobtype) && empty($keywords)){ 

       $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          country = '$location' 
          ORDER BY id_job DESC"; 

       }elseif(isset($location) && isset($jobtype) && empty($keywords)){ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          country = '$location' AND 
          jobType_en = '$jobtype' 
          ORDER BY id_job DESC"; 

       }elseif(isset($location) && isset($jobtype) && isset($keywords)){ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          country = '$location' AND 
          jobType_en = '$jobtype' AND 
          title_en LIKE '%$keywords%' 
          ORDER BY id_job DESC"; 

       }elseif(empty($location) && isset($jobtype) && empty($keywords)){ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          jobType_en = '$jobtype' 
          ORDER BY id_job DESC"; 

       }elseif(empty($location) && isset($jobtype) && isset($keywords)){ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          jobType_en = '$jobtype' AND 
          title_en LIKE '%$keywords%' 
          ORDER BY id_job DESC"; 
       }elseif(empty($location) && isset($jobtype) && isset($keywords)){ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          jobType_en = '$jobtype' AND 
          title_en LIKE '%$keywords%' 
          ORDER BY id_job DESC"; 
       } 
       else{ 
        $sql_jobs = "SELECT * FROM jobs 
          WHERE 
          active = '1' 
          ORDER BY id_job DESC"; 
       } 

       $consultaJob = mysql_query($sql_jobs); 

Answer 1

你可能想尝试的strlen（而不是isset或空

否则你必须同时使用isset()和empty()

empty()如果该变量不存在，则不会生成警告。

strlen（）将在NULL和''上返回零。

您可以通过将零更改为较高值来设置最小长度。

elseif(strlen($location) > 0 && strlen($jobtype) > 0 && strlen($keywords) > 0){ 

如果提交的值是文本输入，那么你可能需要微调也：

elseif(strlen(trim($location)) > 0 && strlen(trim($jobtype)) > 0 && strlen(trim($keywords)) > 0){ 

Answer 2

您需要简化代码：

$sql = "SELECT * FROM jobs 
          WHERE 
          active = '1' AND 
          country = '$location' "; 
$order = "ORDER BY id_job DESC"; 
$where = ""; 

if(isset($jobtype) { $where .= " AND jobType_en = '$jobtype'"; } 
if(isset($keywords) { $where .= " AND title_en LIKE '%$keywords%'" }; 

$sqlx = $sql . $where . $order; 

显然你是处理$ jobtype和$关键字。在这个处理过程中，您应该设置一些默认值，以便稍后处理。

$jobtitle = (isset($_GET['jobtitle'])) ? $_GET['jobtitle'] : ""; 
$keywords= (isset($_GET['keywords'])) ? $_GET['keywords'] : ""; 

这样，您就可以使用：

if($jobtitle != '') { .... }