通过php提交到mysql后js验证

Question

我想从一个html表单提交数据到mysql，但是在我这样做之前我想用一些javascript来验证用户输入。目前发生的事情是，一旦用户点击提交，js验证工作正常，并且调用了php文件，但是当向数据库提交数据时，这些字段都是空的。这是我目前有：

HTML表单

<form action="StudentSubmit.php" action="validating()" method="post" class="fancy-form"> 
. 
. 
<div> 
    <div class="fieldgroup"> 
      <em>*</em><input type="text" name="addressline1" id="addressline1" class="required" minlength="7"> 
      <label for="addressline1">Street #1</label> 
    </div> 
</div> 
. 
. 
. 
<div class="submit-wrap"> 
    <button type="submit" class="submit">Submit</button> 
</div> 

JS

checkdata = 
    { 
     checkNull: function(textboxval) { 
      if ($(textboxval).val() == null || $(textboxval).val() == "") { 
       return false; 
      } 
      else { 
       return true; 
      } 
     }, 
     checkNumeric: function(textboxval) { 
      var value = $(textboxval).val(); 
      if ((parseFloat(value) == parseInt(value)) && !isNaN(value)) { 
       return true; 
      } else { 
       return false; 
      } 
     }, 
     checkEmail: function(textboxval) { 
      var x = $(textboxval).val(); 
      var atpos = x.indexOf("@"); 
      var dotpos = x.lastIndexOf("."); 
      if (atpos < 1 || dotpos < atpos + 2 || dotpos + 2 >= x.length) 
      { 
       return false; 
      } 
     }, 
     isChecked: function(textboxval) { 
      if ($(textboxval).is(':checked') == false) 
      { 
       return false; 
      } else { 
       return true; 
      } 
     } 
    }; 

function validating() { 
var returnValue = true; 


var boxFields = new Array(); 
boxFields[0] = "#firstname"; 
boxFields[1] = "#lastname"; 
boxFields[2] = "#addressline1"; 
boxFields[3] = "#homephone"; 
boxFields[4] = "#cellphone"; 
boxFields[5] = "#city"; 
boxFields[6] = "#state"; 
boxFields[7] = "#guardian"; 
boxFields[8] = "#zip"; 
boxFields[9] = "#emergencycontactperson"; 
boxFields[10] = "#emergencycontactphone"; 

var nullValue = true; 
for (var i = 0; i < boxFields.length; i++) { 
    if (checkdata.checkNull(boxFields[i]) == false) { 
     nullValue = false; 
     break; 
    } 

} 

var numFields = new Array(); 
numFields[0] = "#homephone"; 
numFields[1] = "#cellphone"; 
numFields[2] = "#emergencycontactphone"; 

var numericValue = true; 
for (var j = 0; j < numFields.length; j++) { 
    if (checkdata.checkNumeric(numFields[j]) == false) { 
     numericValue = false; 
     break; 
    } 

} 

var checkFields = new Array(); 
checkFields[0] = "#signature"; 

var checkValue = true; 
for (var k = 0; k < checkFields.length; k++) { 
    if (checkdata.isChecked(checkFields[k]) == false) { 
     checkValue = false; 
     break; 
    } 

} 

if (nullValue == false || numericValue == false || checkValue == false) { 
    returnValue = false; 
    alert("Some required fields were left empty"); 
    return false; 
} else { 
    return true; 
} 
} 

PHP

<?php 
$firstname = $_POST['firstname']; 
$lastname = $_POST['lastname']; 
$dob = $_POST['dob']; 
$emergencycontact = $_POST['emergencycontactperson']; 
$address = $_POST['addressline1'] . " " . $_POST['addressline2']; 
$city = $_POST['city']; 
$state = $_POST['state']; 
$zip = $_POST['zip']; 
$homephone = $_POST['homephone']; 
$cellphone = $_POST['cellphone']; 
$guardian = $_POST['guardian']; 
$inneighborhood = 0; 
if ($zip == "49503") 
    $inneighborhood = 1; 

$con = mysqli_connect("localhost", "cookarts_root", "password_here", "cookarts_database"); 

// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$sql = "INSERT INTO student (FirstName, LastName, DOB, EmergencyContact, Address, City,  State, ZIP, CellPhone, HomePhone, Guardian, InNeighborhood) 
VALUES 
    ('$firstname', '$lastname', '$dob', '$emergencycontact', '$address', '$city', '$state', '$zip', '$cellphone', '$homephone', '$guardian', '$inneighborhood')"; 

if ($con->query($sql) === TRUE) { 
    header('Location: http://cookartscenter.net/ThankYou.php') ; 
} 
else { 
    echo 'Error: '. $con->error; 
} 

mysqli_close($con); 
?> 

只是为了clearify，有连接到数据库没问题，我的问题是在JS验证后，php无法正确从表单中提取数据。如果你需要更多的我的html代码，我很乐意提供它。谢谢。

Answer 1

您正在使用javascript替换表单标记中的表单动作。如果您设置了两个属性，则最后一个属性将是使用的属性。您可以在按钮或onsubmit（）上使用onclick（）。

<form action="StudentSubmit.php" onsubmit="validating()" method="post" class="fancy-form"> 

Javascript验证只适用于使用浏览器的人。它不适用于垃圾邮件或关闭它的人。最好的办法是在服务器端使用PHP验证，然后在将输入注入数据库之前清理输入。

还要检查瓦尔，以确保他们没有空：

if(!empty(strip_tags($_POST['firstname']))){ 
$firstname = strip_tags($_POST['firstname']); 
} 

$firstname = $mysqli->real_escape_string($firstname); 

你也可以报一个PDO连接

$conn->quote($firstname) 

Answer 2

变化

<form action="StudentSubmit.php" action="validating()" method="post" class="fancy-form"> 

到

<form action="StudentSubmit.php" method="post" class="fancy-form"> 

，并添加以下：

$('form').submit(function() 
{ 

    if (!validating()) 
    { 
    return false; 
    } 

} 
)