使用％通配符的Mysqli准备语句

问题描述：

Im am trying to build a dynamic prepared statement so that I can reuse the code. I am running into a problem when using %?% in my prepared state as it used LIKE. My code is as follows:

$where = " First_Name LIKE '%?%' ";
$vals = array('Mike');
$type = 's';

$dbd = mysqli_stmt_init($dbconnection);
if (mysqli_stmt_prepare($dbd, "SELECT * FROM Contacts WHERE $where" )) {
mysqli_stmt_bind_param($dbd, $type, ...$vals);
if (!mysqli_stmt_execute($dbd)) {
echo "Execute Error: " . mysqli_error($dbconnection);
        } else {
            //do nothing
        }
    } else {
        echo "Prep Error: " . mysqli_error($dbconnection);
    }
mysqli_stmt_get_result($dbd);

So when I use "First_Name = ?" it works fine so I think my issue is with the '%?%'. I have searched resolutions but couldn't find anything related to my dynamic prepared statement. Thank you for any help.

我正在尝试构建一个动态预准备语句，以便我可以重用代码。我在准备状态下使用％？％时遇到问题，因为它使用了LIKE。我的代码如下： p>

  $ where =“First_Name LIKE'％？％'”; 
 $ vals = array（'Mike'）; 
 $ type =  's'; 
 
 $ dbd = mysqli_stmt_init（$ dbconnection）; 
if（mysqli_stmt_prepare（$ dbd，“SELECT * FROM Contacts WHERE $ where”））{
mysqli_stmt_bind_param（$ dbd，$ type，... $  vals）; 
if（！mysqli_stmt_execute（$ dbd））{
echo“执行错误：”。  mysqli_error（$ dbconnection）; 
} else {
 //什么都不做
} 
}其他{
 echo“Prep Error：”。  mysqli_error（$ dbconnection）; 
} 
mysqli_stmt_get_result（$ dbd）; 
  code>  pre> 
 
 所以当我使用“First_Name =？”时 它工作正常所以我认为我的问题是'％？％'。 我已经搜索过决议但找不到任何与我的动态准备声明有关的内容。 感谢您的帮助。 p> 
  div>

答

You need to bind the complete value, not just a portion of it. This means doing:

$where = "First_Name LIKE ?"

And then binding:

$vals = array('%Mike%');

使用％通配符的Mysqli准备语句

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