Question

我已经创建了一个将数据导入SQL数据库的表单但是当我填写信息时，我收到了这样的错误：

错误：INSERT INTO andrea22_search（名称，TodayDate，ProjectNumber，Area，DoneBefore，HaveChangesMade，SafeAccess，ElectricalEquipment，MachineGuarding，CorrectEquipment，SDS，ControlToxic，PPE，Hazard，OtherHazard，ControlHazard，MonitorProcess，AdditionalComments）VALUES（'Andi' ，'2017-01-03'，'32a'，'AS'，'是'，'否'，'N / A'，'是'，'否'，'N / A'，'是'，'否”， ''， ''， ''， '是的'， ''）表'andrea22_search.andrea22_search'不存在

如果我查看submit.php文件中的代码，我看不到 andrea22_search.andrea22_search 的来源。这个错误当然会阻止我向数据库添加数据。

$dbname = "andrea22_search";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}


$name = $_POST['input4'];
$date = $_POST['date'];
$projectnumber = $_POST['input1'];
$area   = $_POST['input2'];
$donebefore = $_POST['radio9'];
$changesmade    = $_POST['radio8'];
$safeaccess = $_POST['radio11'];
$electrical = $_POST['radio5'];
$machineguarding    = $_POST['radio6'];
$correctequipment   = $_POST['radio4'];
$sds    = $_POST['radio3'];
$controltoxic   = $_POST['radio1'];
$ppe    = $_POST['checkbox[]'];
$otherhazards   = $_POST['input3'];
$controlofhazards   = $_POST['checkbox1[]'];
$monitor    = $_POST['radio12'];
$comments   = $_POST['input'];

$sql = "INSERT INTO search (Name, TodayDate, ProjectNumber, Area, DoneBefore, HaveChangesMade, SafeAccess, ElectricalEquipment, MachineGuarding, CorrectEquipment, SDS, ControlToxic, PPE, Hazard, OtherHazard, ControlHazard, MonitorProcess, AdditionalComments) VALUES ('$name', '$date','$projectnumber','$area','$donebefore','$changesmade','$safeaccess','$electrical','$machineguarding','$correctequipment','$sds','$controltoxic','$ppe','$otherhazards','$controlofhazards','$monitor','$comments')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

我哪里出错了？

当我尝试从表单输入信息到SQL时，我得到一个表名重复的错误

0 个答案: