Question

我有一个html表单，可以自动选中表单中的所有元素。提交后，它调用第一个元素（order_number），并且光标停留在第二个字段中，这样某人可以使用相同的order_number扫描Multiple（Chip_numbers）。

要重置整个表单，我想刷新页面，然后弹出错误消息，光标自动放入第二个元素。

这是我的剧本：

<?php
$servername = "host";
$username = "username";
$password = "password";
$dbname = "JVSIntranet";

$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT INTO MICROCHIP_TBL (chip_number,order_number)
VALUES 
('$_POST[chip_number]','$_POST[order_number]')";

IF (mysqli_query($conn, $sql)) {
    echo "Great Job! You've entered it correctly!";
} else {
    echo "Error: TRY AGAIN HUMAN!";
}

$value = "";
if( isset( $_POST ["order_number"] )) $value = $_POST ["order_number"];

mysqli_close($conn);
?>

<html>
<head>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<script type="text/javascript">
  function getNextElement(field) {
    var form = field.form;
    for ( var e = 0; e < form.elements.length; e++) {
        if (field == form.elements[e]) {
            break;
        }
    }
    return form.elements[++e % form.elements.length];
}

function tabOnEnter(field, evt) {
if (evt.keyCode === 13) {
        if (evt.preventDefault) {
            evt.preventDefault();
        } else if (evt.stopPropagation) {
            evt.stopPropagation();
        } else {
            evt.returnValue = false;
        }
        getNextElement(field).focus();
        return false;
    } else {
        return true;
    }
}

</script>
</head>

<body onLoad="document.chip_insert.chip_number.focus();">
<center>
<h1>Jeffers HomeAgain Microchip Entry</h1>


<form name="chip_insert"  id="chip_insert" action="<?php echo $PHP_SELF;?>" method="post" >
Order Number: <input tabindex="1" maxlength="12" type="text" name="order_number"  id="order_number" value="<?php echo $value; ?>"  required="required"onkeydown="return tabOnEnter(this,event)"  /><br /><br />
Tag Number: <input tabindex="2" maxlength="15" type="text" name="chip_number" id="chip_number" required="required" /><br /><br />
<input tabindex="7" type="submit" />
</center>
</form>

</body>
</html>

Answer 1

无论何时打开此页面，无论是否有帖子，都会插入数据库。因为没有帖子控制（发布或不控制）下面的代码块：

IF (mysqli_query($conn, $sql)) {
 echo "Great Job! You've entered it correctly!";
} else {
 echo "Error: TRY AGAIN HUMAN!";
}

这部分应该是这样的：

if(isset($_POST)){
  IF (mysqli_query($conn, $sql)) {
    echo "Great Job! You've entered it correctly!";
  } else {
     echo "Error: TRY AGAIN HUMAN!";
 }
}

Answer 2

你的php部分应该是这样的

<?php
$servername = "host";
$username = "username";
$password = "password";
$dbname = "JVSIntranet";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}



$value = "";
if (isset($_POST ["order_number"])) {
    $value = $_POST ["order_number"];
    $sql = "INSERT INTO MICROCHIP_TBL (chip_number,order_number)
VALUES 
('$_POST[chip_number]','$_POST[order_number]')";

    if (mysqli_query($conn, $sql)) {
        echo "Great Job! You've entered it correctly!";
    } else {
        echo "Error: TRY AGAIN HUMAN!";
    }
}


mysqli_close($conn);
?>

并且html </center>应该在</form>标记之后，尽管它不是大问题

Answer 3

此代码还有一些问题......

IF (mysqli_query($conn, $sql)) {
 echo "Great Job! You've entered it correctly!";
} else {
 echo "Error: TRY AGAIN HUMAN!";
}

应该像

if(isset($_POST['chip_number'] && $_POST['chip_number'] != '')
{
  IF (mysqli_query($conn, $sql)) {
   echo "Great Job! You've entered it correctly!";
  } else {
   echo "Error: TRY AGAIN HUMAN!";
  }
}

和

$sql = "INSERT INTO MICROCHIP_TBL (chip_number,order_number)
VALUES 
('$_POST[chip_number]','$_POST[order_number]')";

应该有点像

$sql = "INSERT INTO MICROCHIP_TBL (chip_number,order_number)
VALUES 
('{$_POST['chip_number']}','{$_POST['order_number']}')";

甚至最后一件事都不是正确的类型来称呼它...... 你可以查看http://php.net/manual/en/mysqli-stmt.bind-param.php

实际上会是什么样的

$stmt = mysqli_prepare($link, "INSERT INTO MICROCHIP_TBL VALUES (?, ?)");
mysqli_stmt_bind_param($stmt, 'dd', $_POST['chip_number'], $_POST['order_number']);
//....
mysqli_execute();

或

$stmt = mysqli_prepare($link, "INSERT INTO MICROCHIP_TBL SET chip_number=?,order_number=?");
mysqli_stmt_bind_param($stmt, 'dd', $_POST['chip_number'], $_POST['order_number']);
//....
mysqli_execute();

这些样本不是像$ db = new mysqli（...）这样的对象构建的;但你应该考虑以这种方式使用它：）

为什么我的表单会在页面加载时自动提交？

3 个答案: