我是php的初学者。 我有一个像这样的数据表
| 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 |
我有一个像这样的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>
现在我想要基于多重选择的查询
如果在所有三个下拉列表中选择“any-”,则应该获取所有表格行
如果选择了任何两个项目,即location = india和profession = designer,那么它应该只获取第一行和第七行,其中两个选择值都匹配
请帮助我使用php根据表单选择值
获取结果这是我的php检索帖子值
<?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post值进行过滤。请帮忙
我试过下面的一个。我可以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'] ."'";
答案 0 :(得分:0)
尝试以下代码
$where = '';
if(isset($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 tablename $where";
答案 1 :(得分:0)
// 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 $sql; // SELECT * FROM tablename WHERE source = 'from x' AND profession = 'designer';
exit;
答案 2 :(得分:0)
<?php
if(isset($_POST['submit']))
{
mysql_connect("localhost","root","");
mysql_select_db("alldata");
$select='SELECT serial,name,email,phone,location,profession,source From <table_name> ';
$where =' Where 1 ';
if(isset($_POST['location']) && trim($_POST['location'])!='')
{
$where.=' and location= '.mysql_real_escape_string($_POST['location']);
}
if(isset($_POST['source']) && trim($_POST['source'])!='')
{
$where.=' and source= '.mysql_real_escape_string($_POST['source']);
}
if(isset($_POST['profession']) && trim($_POST['profession'])!='')
{
$where.=' and profession= '.mysql_real_escape_string($_POST['profession']);
}
$query=$select.$where;
$request=mysql_query($query) or die('query_error');
echo "<table border='1'>";
echo "<tr><td>serial</td><td>name</td><td>email</td><td>phone</td><td>location</td><td>profession</td><td>source</td></tr>";
while($result=mysql_fecth_array($request))
{
echo "<tr>";
echo "<td>".$result['serial']."</td>";
echo "<td>".$result['name']."</td>";
echo "<td>".$result['email']."</td>";
echo "<td>".$result['phone']."</td>";
echo "<td>".$result['location']."</td>";
echo "<td>".$result['profession']."</td>";
echo "<td>".$result['source']."</td>";
echo "</tr>";
}
echo "</table>";
}
?>
<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>