Question

我有一个包含2个选择字段和一个输入的搜索表单，总共3个选项，所以我创建了一些if语句，根据每个字段设置它有自己的查询，但它不能正常工作，都是错误的，结果不对，它与陈述查询混在一起，它没有做对。

以下是搜索表单结果的代码：

$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strlen（）而不是isset或empty

否则您必须同时使用isset()和empty()

如果变量不存在，

empty()不会生成警告。

strlen（）将在NULL和＆＃39;＆＃39;。

上返回零

您可以通过将零值更改为更高的值来设置最小长度。

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

如果提交的值是文本输入，那么您可能还需要修剪：

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;

你显然是在早些时候处理$ jobtype和$ keywords。在此处理过程中，您应该设置一些默认值，以便以后处理更容易。

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

这样你可以使用：

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

搜索表格多输入

2 个答案: