Question

当我单独拉出每个leadstatus（？leadstatus = New，？leadstatus = Hot等）时，他们会工作，但是当我试图获得All时，我似乎无法让它工作。您可以看到页面上的默认值是新的潜在客户。

`$query = "SELECT * FROM contacts WHERE contacttype IN ('New','Buyer','Seller','Buyer / Seller','Investor') AND leadstatus = 'New' ORDER BY date DESC";

    if(isset($_GET['leadstatus']) && in_array($_GET['leadstatus'], array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed')))
    {      
    $status = $_GET['leadstatus'];   
    $query = "SELECT * FROM contacts WHERE leadstatus = '".$status."' ORDER BY contacts.date DESC";  
    }`

以下是我尝试过的一些没有运气的字符串：

?leadstatus=New&leadstatus=Hot&leadstatus=Warm&leadstatus=Rejected&leadstatus=Cold - Only pulls last listed, which is Cold

?leadstatus[]=New&leadstatus=[]Hot&leadstatus[]=Warm&leadstatus[]=Rejected&leadstatus[]=Cold - Returns default, which is New

?leadstatus=New&Hot&Warm&Rejected&Cold

返回默认值，即New

Answer 1

if(isset($_GET['leadstatus']) && $_GET['leadstatus'] == "all") {
    $query = "SELECT * FROM contacts ORDER BY contacts.date DESC";  
} else if (in_array($_GET['leadstatus'], array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed'))) {      
    $status = $_GET['leadstatus'];   
    $query = "SELECT * FROM contacts WHERE leadstatus = '".$status."' ORDER BY contacts.date DESC";  
}

然后，make leadstatus = all。

Answer 2

试试这个：

if(isset($_GET['leadstatus']) && in_array($_GET['leadstatus'], array('New', 'Hot', 'Warm', 'Cold', 'Rejected', 'Closed')))
{      
  $status = $_GET['leadstatus'];   
  if(!empty($status)) {
    $query = "SELECT * FROM contacts WHERE leadstatus = '".$status."' ORDER BY contacts.date DESC";  
  } else {
    $query = "SELECT * FROM contacts ORDER BY contacts.date DESC"; 
  }
}`

但是，我是否还建议您使用参数化查询？你在这里对SQL注入攻击持开放态度。

Answer 3

这样的事情应该符合多个条件，允许你一次混合和匹配几个，而不是1个或全部。

$status = join(',',$_GET['leadstatus']); 
$query = "SELECT * FROM contacts WHERE leadstatus IN($status) ORDER BY contacts.date DESC";

查询表中所有记录的字符串

3 个答案: