Question

我有一个包含3个元素的搜索表单关键词，行业（下拉）＆amp;位置

我正在尝试创建搜索功能，如果选择了所有3个元素或基于单个值或者提交了$ _POST数据的任何元素的组合，将查询数据库

它只是不起作用

我的查询代码在

之下

        if(isset($_POST['keywords']))
    {
    $keywords = $_POST['keywords'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE (`title` LIKE '%$keywords%' OR `description` LIKE '%$keywords%') LIMIT 0,50");
    $resultsfor = $_POST['keywords'];
    }
    elseif(isset($_POST['keywords']) && isset($_POST['location']))
    {
    $keywords = $_POST['keywords'];
    $location = $_POST['location'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE (`title` LIKE '%$keywords%' OR `description` LIKE '%$keywords%') AND (`location` LIKE '%$location%') LIMIT 0,50");
    $resultsfor = $keywords .' jobs in '.$location;
    }
    elseif(isset($_POST['keywords']) && isset($_POST['location']) && !empty($_POST['industry']))
    {
    $keywords = $_POST['keywords'];
    $location = $_POST['location'];
    $catno = $_POST['industry'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE (`title` LIKE '%$keywords%' OR `description` LIKE '%$keywords%') AND (`location` LIKE '%$location%') AND (`catno` = '$catno') LIMIT 0,50");
    $resultsfor = $keywords .' jobs in '.$location;
    }
    elseif(isset($_POST['industry']) && empty($_POST['location']))
    {
    $industry = $_POST['industry'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE `catno` = '$industry' LIMIT 0,50");
    $resultsfor = $_POST['industry']. ' jobs';
    }
    elseif(isset($_POST['industry'])&& isset($_POST['location']))
    {
    $industry = $_POST['industry'];
    $location = $_POST['location'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE (`catno` = '$industry') AND (`location` LIKE '%$location%') LIMIT 0,50");
    $resultsfor = $_POST['industry']. ' jobs in '.$location;
    }
    elseif(isset($_POST['location']))
    {
    $location = $_POST['location'];
    $viewq = mysqli_query($con,"SELECT * FROM `listings` WHERE (`location` LIKE '%$location%') LIMIT 0,50");
    $resultsfor = $_POST['location']. ' jobs';
    }

    $view = mysqli_fetch_assoc($viewq);

非常感谢一些帮助，我花了很多时间在这上面提前致谢

**它确实提取数据虽然不正确，如果我搜索仓库（留下行业下拉空白）但设置位置到莱斯特仍然查询所有结果而不是莱斯特如果我从下拉菜单中选择行业并将其他字段留空，我将获得所有类别中的所有作业，而不仅仅是所选类别

让自己感到困惑

Answer 1

我想你的问题是由于你构建if-else条件的复杂方式。一个更简单的方法是按照以下方式进行...

// Get the input variables
$keywords = (isset($_POST['keywords']) && strlen($_POST['keywords']) > 0) ? $_POST['keywords'] : null;
$location = (isset($_POST['location']) && strlen($_POST['location']) > 0) ? $_POST['location'] : null;
$catno = (isset($_POST['industry']) && strlen($_POST['industry']) > 0) ? $_POST['industry'] : null;

$whereUsed = false;
$whereString = "";
$resultsfor = "";

// Add to the WHERE clause if keywords exists.
if ($keywords !== null) {
    if (! $whereUsed) {
        $whereString .= 'WHERE ';
        $whereUsed = true;
    } else {
        $whereString .= ' AND ';
    }
    $whereString .= "(title LIKE '%{$keywords}%' OR description LIKE '%{$description}%')";
    if ($catno === null) {
        $resultsfor .= $keywords;
    }
}
// Add to the WHERE clause if catno exists.
if ($catno !== null) {
    if (! $whereUsed) {
        $whereString .= 'WHERE ';
        $whereUsed = true;
    } else {
        $whereString .= ' AND ';
    }
    $whereString .= "(catno = '{$catno}')";
    $resultsfor .= $catno;
}
// Add to the WHERE clause if location exists.
if ($location !== null) {
    if (! $whereUsed) {
        $whereString .= 'WHERE ';
        $whereUsed = true;
    } else {
        $whereString .= ' AND ';
    }
    $whereString .= "(location LIKE '%{$location}%')";
    if ($catno === null && $keywords === null) {
        $resultsfor = "{$location} jobs";
    } else {
        $resultsfor .= " jobs in {$location}";
    }
}

// Build the SQL query using the WHERE clauses we've built up
$sqlQuery = "SELECT * FROM listings {$whereString} LIMIT 0, 50";

// Execute the query and fetch the response
$viewq = mysqli_query($con, $sqlQuery);
$view = mysqli_fetch_assoc($viewq);

您的原始代码中不需要if-else-if格式。您只是根据是否设置了变量来添加WHERE子句......所以您应该这样做。

请注意，在您的代码中，在我的示例中，SQL查询容易受到SQL注入攻击。我强烈建议调查prepared statements。

Answer 2

我在你的if语句中看到了一些错误。你应该测试从大多数特定到低限制。例如，您的第一个elseif将永远不会发生，因为在这种情况下，第一个if总是会出现。

Answer 3

应该有九种组合来过滤您的查询。你的条件应该是

<div class="tp-cont" ng-style="{ 'background-image': 'url({{bgUrl}})' }">

Answer 4

  if (isset($_POST['keywords']) && isset($_POST['location']) && 
     isset($_POST['industry'])) {
     $keywords = $_POST['keywords'];
     $location = $_POST['location'];
     $catno = $_POST['industry'];
     $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE (`title` LIKE '%$keywords%' OR `description` LIKE '%$keywords%') AND (`location` LIKE '%$location%') AND (`catno` = '$catno') LIMIT 0,50");
     $resultsfor = $keywords . ' jobs in ' . $location;
  } elseif (isset($_POST['keywords']) && isset($_POST['location']) && 
    empty($_POST['industry'])) {
           $keywords = $_POST['keywords'];
      $location = $_POST['location'];
      $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE (`title` 
     LIKE '%$keywords%' OR `description` LIKE '%$keywords%') AND (`location` 
      LIKE '%$location%') LIMIT 0,50");
    $resultsfor = $keywords . ' jobs in ' . $location;
    } elseif (isset($_POST['keywords']) && empty($_POST['location']) && 
     empty($_POST['industry'])) {
      $keywords = $_POST['keywords'];
      $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE (`title` 
      LIKE '%$keywords%' OR `description` LIKE '%$keywords%') LIMIT 0,50");
      $resultsfor = $_POST['keywords'];
   } elseif (empty($_POST['keywords']) && isset($_POST['industry']) && 
   empty($_POST['location'])) {
     $industry = $_POST['industry'];
     $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE `catno` = 
    '$industry' LIMIT 0,50");
     $resultsfor = $_POST['industry'] . ' jobs';
  } elseif (empty($_POST['keywords']) && isset($_POST['industry']) && 
     isset($_POST['location'])) {
     $industry = $_POST['industry'];
       $location = $_POST['location'];
     $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE (`catno` = 
     '$industry') AND (`location` LIKE '%$location%') LIMIT 0,50");
    $resultsfor = $_POST['industry'] . ' jobs in ' . $location;
} elseif (empty($_POST['keywords']) && empty($_POST['industry']) && 
isset($_POST['location'])) {
       $location = $_POST['location'];
           $viewq = mysqli_query($con, "SELECT * FROM `listings` WHERE 
     (`location` LIKE '%$location%') LIMIT 0,50");
      $resultsfor = $_POST['location'] . ' jobs';
   }

     $view = mysqli_fetch_assoc($viewq);

请试试这个

Answer 5

我尝试最小化您的代码并检查它并告诉它是否正常

<?php

    //Enter your code here, enjoy!
    $query          = "SELECT * FROM listings ";
    $eQuery         = "";
    $resultsfor     = "";

    if(isset($_POST['keywords']) && !empty($_POST['keywords']){
        $eQuery     .= " (title LIKE '%".addslashes($_POST['keywords'])."%' OR description LIKE '%".addslashes($_POST['keywords'])."%') ";
        $resultsfor .= $_POST['keywords'];
    }

    if(isset($_POST['location']) && !empty($_POST['location']){
        $eQuery     .=  " (`location` LIKE '%".addslashes($_POST['location'])."%') ";
        $resultsfor .= ' jobs in '.$location;
    }

    if(isset($_POST['industry']) && !empty($_POST['industry'])){
        $eQuery     .= " (`catno` = '".$_POST['industry']."')";
        $resultsfor = (isset($_POST['keywords']) && (isset($_POST['location'])))? $_POST['keywords'] .' jobs in '.$_POST['location'] ;
    }

    $query = mysql_query($con, $query.$eQuery." LIMIT 0,50");

    $view = mysqli_fetch_assoc($query);

elseif搜索查询无效

5 个答案: