我遇到了将LIKE与Wildcard绑定到我在
MySQLi中准备好的语句的问题.我尝试了以下两种方法,如图所示& concat.(用@fancyPants输入更新)
>有没有办法让我可以在绑定发生后查看自己的sql语句？
>如何正确绑定它以获得我想要的结果？
它没有LIKE语句.
- $str = $_POST["searchstr"];
- if(isset($_POST['submit']))
- {
- $price=$_POST['price'];
- if(!empty($_POST['chkbx']))
- {
- foreach($_POST['chkbx'] as $selected)
- {
- $sql= 'SELECT bookTitle,bookPrice FROM nbc_book WHERE catID LIKE "%'.$selected.'%" AND bookTitle LIKE "%'.$str.'%" AND bookPrice < ?';
- $stmt=MysqLi_prepare($con,$sql);
- MysqLi_stmt_bind_param($stmt,"i",$price);
- MysqLi_stmt_execute($stmt);
- MysqLi_stmt_bind_result($stmt,$bookTitle,$bookPrice);
- while ($stmt->fetch()) {
- echo $bookTitle.$bookPrice."<br>";
- }
- }
- }
- }
- $searchStr = 'oracle';
- $sql= 'SELECT bookTitle,bookPrice FROM nbc_book WHERE catID LIKE ? AND bookTitle LIKE "%'.$searchStr.'%" AND bookPrice < ?';
- $stmt=MysqLi_prepare($con,$sql);
- MysqLi_stmt_bind_param($stmt,"ssi",$selected,$price);
- MysqLi_stmt_execute($stmt);
- MysqLi_stmt_bind_result($stmt,$bookPrice);
- while ($stmt->fetch()) {
- echo $bookTitle;
- }
