0

我正在尝试使用 mysqli 实现我的第一个准备好的语句。

目前我有这个：

<?php
$con = new mysqli('example.com', 'user', 'password', 'database');
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();}

$first = $_GET['firstname'];
$last = $_GET['surname'];
$dob = $_GET['dob'];
$address = $_GET['homeaddress'];
$college = $_GET['college'];
$emergname = $_GET['emergencyname'];
$emergnumber = $_GET['emergencynumber'];
$condition = $_GET['condition'];
$conditiondetails = $_GET['conditiondetails'];
$medication = $_GET['medication'];
$medicationdetails = $_GET['medicationdetails'];

if($stmt = $con->prepare("INSERT INTO medical ('forename', 'surname', 'dob', 'address', 'college', 'emergency_name', 'emergency_number', 'condition', 'condition_details', 'medication', 'medication_details') VALUES (:forename, :surname, :dob, :address, :college, :emergencyname, :emergencynumber, :condition, :conditiondetails, :medication, :medicationdetails)")){

$stmt->bind_param(':forename', $first);
$stmt->bind_param(':surname', $last);
$stmt->bind_param(':dob', $dob);
$stmt->bind_param(':address', $address);
$stmt->bind_param(':college', $college);
$stmt->bind_param(':emergencyname', $emergname);
$stmt->bind_param(':emergencynumber', $emergnumber);
$stmt->bind_param(':condition', $condition);
$stmt->bind_param(':conditiondetails', $conditiondetails);
$stmt->bind_param(':medication', $medication);
$stmt->bind_param(':medicationdetails', $medicationdetails);
$stmt->execute();
$stmt->close();} ?>

我之前尝试过使用以下方法进行更改：

<?php
$stmt = $con->prepare("INSERT INTO medical ('forename', 'surname', 'dob', 'address', 'college', 'emergency_name', 'emergency_number', 'condition', 'condition_details', 'medication', 'medication_details') VALUES (?,?,?,?,?,?,?,?,?,?,?)")

$stmt->bind_param('sssssssssss', $first...);
?>

在这两种情况下，我都会收到一条错误消息，指出 $stmt 变量不存在。

关于我哪里出错的任何建议？

4

2 回答 2

2

列名应该用反引号转义，而不是单引号。此外，您不能使用命名参数绑定。

尝试

$stmt = $con->prepare("INSERT INTO medical (`forename`, `surname`, `dob`, `address`, `college`, `emergency_name`, `emergency_number`, `condition`, `condition_details`, `medication`, `medication_details`) VALUES (?,?,?,?,?,?,?,?,?,?,?)")

if (!$stmt)
{
  echo $con->error;
}
于 2012-10-12T20:04:31.873 回答
0

也许您应该尝试使用以下 SQL 语法：

$stmt = $con->prepare("INSERT INTO medical VALUES (:forename, :surname, :dob, :address, :college, :emergencyname, :emergencynumber, :condition, :conditiondetails, :medication, :medicationdetails)");

$stmt->bind_param(':forename', $first);
$stmt->bind_param(':surname', $last);
$stmt->bind_param(':dob', $dob);
$stmt->bind_param(':address', $address);
$stmt->bind_param(':college', $college);
$stmt->bind_param(':emergencyname', $emergname);
$stmt->bind_param(':emergencynumber', $emergnumber);
$stmt->bind_param(':condition', $condition);
$stmt->bind_param(':conditiondetails', $conditiondetails);
$stmt->bind_param(':medication', $medication);
$stmt->bind_param(':medicationdetails', $medicationdetails);
$stmt->execute();

或尝试：

$stmt = $con->prepare("INSERT INTO medical VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)");

$stmt->bind_param('sssssssssss', $first, $second, $third, $fourth, $fifth, $sixth, $seventh, $eighth, $ninth, $tenth, $eleventh);

意思是，在您的 INSERT 语句中没有 Use-Fields 声明。

于 2012-10-12T20:01:46.687 回答