sql语句中使用变量后，结果不能正常读取？

Question

我用变量代替了sql中的语句就无法从mysql数据库中读取先关的数据了？究竟是为什么？

<?php 
    if($_GET["dish_kind_breakfast1"]!="暂无"){
    $sql="SELECT * FROM dishList_table WHERE dish_kind= '".$_GET["dish_kind_breakfast1"]."' AND vegetarian='".$_GET["vegetarian_breakfast1"]."'AND ".$bodyRequireOne."AND" .$bodyRequireTwo;
    $result=$conn->query($sql);
    if($result->num_rows>0){
      $row=$result->fetch_assoc();
        }else{
            echo "there is no result.sorry";
        }
}   
    ?>

代码如果像上面这样写的话，就无法从数据库中读取。虽然 echo $sql 如下：
SELECT * FROM dishList_table WHERE dish_kind= '粥' AND vegetarian='素食'AND yang_deficiency_constitution='TRUE'ANDbalanced_constitution='TRUE'

但如果我把变量$bodyRequireOne和变量$bodyRequireTwo对应的变量直接写入如下，又可以获取数据：

<?php 
    if($_GET["dish_kind_breakfast1"]!="暂无"){
           $sql="SELECT * FROM dishList_table WHERE dish_kind= '".$_GET["dish_kind_breakfast1"]."' AND vegetarian='".$_GET["vegetarian_breakfast1"]."'AND yang_deficiency_constitution='TRUE'AND balanced_constitution='TRUE'";
       $result=$conn->query($sql);
         if($result->num_rows>0){
            $row=$result->fetch_assoc();
        }else{
            echo "there is no result.sorry";
        }

}   
    ?>

我把 $sql echo 出来对比了，也没有发现什么错误，究竟为什么？还望指点，谢谢！

Answer 1

看起来没有啥错误，使用转义试一试：

$bodyRequireOne = mysqli_real_escape_string($conn, $bodyRequireOne);
$bodyRequireTwo = mysqli_real_escape_string($conn, $bodyRequireTwo);
$sql = "SELECT * FROM dishList_table WHERE dish_kind = '".$_GET["dish_kind_breakfast1"]."' 
          AND vegetarian = '".$_GET["vegetarian_breakfast1"]."' 
         AND ".$bodyRequireOne." AND ".$bodyRequireTwo;