Question

自动生成数据绑定＆＃34;没有数据提供参数＆＃34;

我创建了一个脚本来查找表单发送的所有$_POST值，创建查询并绑定数据以通过MySQLi提交

它一直运行到最后并导致错误消息：

＆＃34;没有为准备好的声明和＃34;
中的参数提供数据

这让我感到难过，因为我的echo语句似乎表明数据已被传递和处理。

<?php
$user = "user";
$pw = "pw";
$url = "myURL"; 
$db = "myDB";
/*** create a new mysqli object with default database***/
$mysqli = @new mysqli($url, $user, $pw, $db);

/* check connection */ 
if(!mysqli_connect_errno())
    {
    /*** if we are successful ***/
    echo 'Connected Successfully<br />';

    /*** Create the SQL statement ***/
   /***Put each variable on separate line so I can use spreadsheet
    this helps make sure the number of values is equivalent in all parts ****/
    $type = "'";
    $params = "";
    $valholder = "";
    /*** Automatically read the variables sent from study software and generate the MySQLi binding */
    foreach ($_POST as $param_name => $param_val) {
    if(isset($_POST[$param_name])){
    if($param_val==='Submit'){
    }
    else{
    if(is_numeric($param_val)){
    $type.="i";
    $vals.=$param_val.",";
    }
    else{
    $type.="s";
    $vals.="'".$param_val."',";
    }
    //echo "Param: $param_name; Value: $param_val<br />\n";
    $valholder.="?,";
    $params.=$param_name.",";
    }
    }

    }


    $vals = rtrim($vals, ",");
    $params = rtrim($params, ",");
    $valholder = rtrim($valholder, ",");
    $type.= "'";
    echo $type.'<br />';
    echo $vals.'<br />';
    /*** Initialize a statement object for mysqli_stmt_prepare() ***/
    $stmt = mysqli_stmt_init($mysqli);
    /*** prepare the statement ***/
    $sql = 'INSERT INTO myTable ('.$params.') VALUES('.$valholder.')';
    echo $sql.'<br/>';

    if (mysqli_stmt_prepare($stmt, $sql))
        {
        echo 'mysqli successfully prepared<br/>';
        /*** bind the parameters ***/
        }
        else{
        //die( 'stmt error: '.mysqli_stmt_error($stmt) );
        Echo 'error with binding <br />';
        }

        mysqli_stmt_bind_param($stmt,$type,$vals);

        /*** set parameters and execute ***/
    if ( !mysqli_execute($stmt) ) {
          die( 'stmt error: '.mysqli_stmt_error($stmt) );
        }   

   }
else
    {
    /*** if we are unable to connect ***/
    echo 'Unable to connect';
    exit();
    }
?>

回声输出为：

已成功连接
＆＃39; SSI＆＃39;
＆＃39;评估1＆＃39;，＆＃39;答案1＆＃39;，111222333
INSERT INTO myTable （评估，CorrectAnswers，CRN）价值观（？，？，？）
mysqli成功
准备好的stmt错误：没有为准备好的参数提供数据声明

我错过了什么？

没有为准备好的声明中的参数提供数据

0 个答案: