Question

我在使用php作为输入和输出开发高级搜索代码时遇到了问题，sql用于选择和过滤数据..

php代码：

<form action="index.php?Type=Advance" method="post">
      <input type="text"   name="name">
                <input type="text"   name="sponsor">

    <select size="1" name="gender" id="">
    <option value="male">male</option>
    <option value="female">female</femal>
</select>
                       <select size="1" name="address" id="">
    <option value="x">x</option>
    <option value="y">y</option>
    <option value="z">z</option>
       </select>
             <input type="submit">
  </form>

然后我宣布变量

公共职能AdvanceSearch（$ name，$ sponsor，$ gender，$ address） {

$cheack = "";
if(isset($name)&&$name != ""){
    $cheack.="  name =  '$name' ";
}
if(isset($sponsor)&&$sponsor != ""){
    $cheack.=" AND sponsor =  '$sponsor' ";
}
if(isset($gender)&&$gender != ""){
    $cheack.=" AND gender =  '$gender' ";
}
if(isset($address) &&$address != "" ){
    $cheack.=" AND workplace =  '$address' ";
}
   $DB = mysql_query("SELECT * FROM table WHERE 1 =  1 ".$cheack);
   echo "SELECT * FROM user WHERE ".$WHQ;
   exit();

实际上它可以工作，但是如果我没有插入名称...... sql语句将是这样的

SELECT * 
FROM table 
WHERE AND sponsor = 'www' 
    AND gender = 'male'

如果我想搜索表但没有插入名称怎么办...如果我没有插入名称，我怎么能让sql语句有效。

Answer 1

对此的典型解决方案是始终首先添加真实条件，例如1=1。没有任何额外条件的查询将成为

SELECT * FROM table WHERE 1=1

当你添加任何AND条件时，你可以将它们添加到最后，没有特殊情况的第一个或最后一个条件：

SELECT * FROM table WHERE 1=1 AND sponsor = 'www' AND gender = 'male'

请注意，如果您使用OR代替AND，则第一个条件应为false，例如0=1。

Answer 2

您可以使用以下标志变量：

$cheack = "";
$flag = False;

if(isset($name)&&$name != ""){
 $cheack.="  name =  '$name' ";
 $flag =True;
}

if(isset($sponsor)&&$sponsor != ""){
 if($flag){
  $cheack.="AND ";
 }
 $cheack.="sponsor =  '$sponsor' ";
}

if(isset($gender)&&$gender != ""){
 if($flag){
  $cheack.="AND ";
 }
 $cheack.="gender =  '$gender' ";
}

if(isset($address) &&$address != "" ){
 if($flag){
  $cheack.="AND ";
 }
 $cheack.="workplace =  '$address' ";
}

php / mysql高级搜索页面

2 个答案: