2016-10-03 61 views
-3
include 'connect.php'; 
msql_select_db("u972015033_jobss"); or die("Could not find DB"); 
if(isset($_POST['search'])){ 
    $searchq = $_POST['search']; 
    $query = msql_query("SELECT * FROM job_search WHERE job_name LIKE '%$searchq%'"); or die("Could not find"); 
    $count = mysql_num_rows($query); 
    if($count == 0){ 
     $output = 'There was no results found'; 
     else{ 
      while($row = mysql_fetch_array($query)){ 
       $jobname = $row['job_name']; 
       $jobdesc = $row['job_desc']; 
       $jobcomp = $row['job_company']; 
       $output .= '<div> '.$jobname.' '.$jobdesc.' '.$jobcomp.'</div>'; 
       echo = "$output"; 
      } 
     } 
    } 
} 

我不知道什么是请帮助 它的一个基本的脚本来查询数据库,并显示 输出简单的PHP错误T_print错误

+0

您没有给出mysql_select_db()中的链接标识符 –

+0

您是否缺少关闭} if if before else。 –

+2

您的代码中存在多个语法错误,请先进行一些基本的调试。 – Epodax

回答

0

这里的问题是错误的清单:

  1. msql_select_db(“u972015033_jobss”);或死(“找不到数据库”); //它应该是mysql_select_db(“u972015033_jobss”)或die(“Could not find DB”);
  2. $ query = msql_query(“SELECT * FROM job_search WHERE job_name LIKE'%$ searchq%'”);或死(“找不到”); //必须是:$ query = mysql_query(“SELECT * FROM job_search WHERE job_name LIKE'%$ searchq%'”)或死(“Could not find”);
  3. before else close'}'
  4. echo =“$ output”; //删除“=”
+0

基本上我说的也是,但有一点小差别:我不认为他错过了大括号,他把它放在错误的地方,在整个街区之后 – mishu

+0

@mishu,是的你是对的。我没有注意到你的评论 –