php - 使用 HTML 表单作为方案动态查询数据库

Question

我是 php 的初学者。我有一个类似这样的数据表

| serial    | | name            | | email           | | phone       | | location    | | profession  | | source  |
-----------------------------------------------------------------------------------------------------------------------------
| 1         | | Mr first        | | a@example.com   | | 123456780   | | India       | | Designer    | | From X  |
| 2         | | Mr second       | | b@example.com   | | 123456781   | | US          | | Designer    | | From Y  |
| 3         | | Mr third        | | c@example.com   | | 123456782   | | US          | | Engineer    | | From X  |
| 4         | | Mr fourth       | | d@example.com   | | 123456783   | | US          | | Disigner    | | From Z  |
| 5         | | Mr fifth        | | e@example.com   | | 123456784   | | India       | | Engineer    | | From Y  |
| 6         | | Mr sixth        | | f@example.com   | | 123456785   | | UK          | | Designer    | | From X  |
| 7         | | Mr seventh      | | g@example.com   | | 123456786   | | India       | | Designer    | | From X  |

我有一个这样的 html 表单

<form method="post" action="search.php">
 <select name="location">
 <option value="" selected="selected">-any-</option>
 <option value="UK">UK</option>
 <option value="India">India</option>
 <option value="US">US</option>
 </select>

 <select name="source">
 <option value="" selected="selected">-any-</option>
 <option value="From X">From X</option>
 <option value="From Y">From Y</option>
 <option value="From Z">From Z</option>
 </select>

 <select name="profession">
 <option value="" selected="selected">-any-</option>
 <option value="Designer">Designer</option>
 <option value="Engineer">Engineer</option>
    </select>
 <input type="submit" value="submit">
</form>

现在我想要基于多选的查询

1. 如果在所有三个下拉列表中都选择了“any-”，它应该获取所有表格行

2. 如果选择了任何两个项目，即 location=india 和 profession=designer，那么它应该只获取两个选择值都匹配的第 1 行和第 7 行

请帮助我使用 php 获取基于表单选择值的结果

这是我用来检索帖子值的 php

<?php

mysql_connect("localhost","root","");
mysql_select_db("alldata");

if(isset($_POST['submit'])) {
$source=$_POST['source'];
$profession=$_POST['profession'];
$location=$_POST['location'];
}

?>

我的 php 文件现在看起来像

    <?php 
$conn = mysql_connect ("localhost", "root", "") or die ('I cannot connect to the database because: ' . mysql_error()); 
$selected = mysql_select_db ("alldata") 
or die ("Could not select database because: " . mysql_error()); 

if(isset($_POST['submit'])) {
$source=$_POST['source'];
$profession=$_POST['profession'];
$location=$_POST['location'];
}
$where = '';
if(isset($location) && !empty($location)){
    $where .= "location ='$location' AND ";
}
if(isset($profession) && !empty($profession)){
     $where .= "profession ='$profession' AND ";
}
if(isset($source) && !empty($source)){
        $where .= "source ='$source' AND ";
}
$where = substr($where, 0, (strlen($where) - 4));
$where = ($where != '') ? "WHERE $where":'';
$sql= "select * from data $where";
$result = mysql_query($sql,$conn)or die (mysql_error()); 
if (mysql_num_rows($result)==0){ 
echo "No Match Found"; 
}else{ 
while ($row = mysql_fetch_array($result)){ 
echo "" .$row['name']." " .$row['email']." ".$row["phone"]." ".$row["source"]." ".$row["profession"]." ".$row["location"]."<br>";

echo "<br>"; 
echo "---------------------------------------------------------------------"."<br>"; 
} 
} 
mysql_close(); 
?>

它获取所有行而不是基于发布值进行过滤。请帮忙

我尝试了下面的一个。我能够通过过滤值来获得结果。但是，如果有任何空选择，我需要从过滤中跳过该值。这是我的代码

$sql = "select * from data
        where location = '".$_POST['location']."' 
        AND profession = '". $_POST['profession'] ."' 
        AND source = '". $_POST['source'] ."'";

Answer 1

// get your real values from $_POST array here 
$location   = null; // i.e. "any"
$source     = 'from x';
$profession = 'designer'; 

$filters = array(
    'location'   => $location, 
    'source'     => $source,
    'profession' => $profession,
    // etc
);

$where = 'WHERE';

$sql = "SELECT * FROM tablename";

foreach ($filters as $field => $value) {
    if($value) {
        $sql .= " $where $field = '$value'";
        $where = 'AND';
    }
}

$sql .= ";";

echo $sql; // SELECT * FROM tablename WHERE source = 'from x' AND profession = 'designer';
exit;