将HTML表单中的输入数据插入JSON对象，然后将其存储在MYSQL中

时间：2014-10-07 06:06:17

标签： php html mysql ajax json

我正在尝试从HTML中获取用户注册表单中的数据，然后将数据推送到JSON，然后获取JSON并存储到MySQL中。请帮我。

HTML 

    <form id="myForm" action="userInfo.php" method="post">
        <table align="center">
            <tr>
                <td><label for="FirstNameLabel" class="tableproperties">First Name</label></td>
                <td><input type="text" class="signupTextBoxStyle" name="firstName" placeholder="Enter First Name" id="FirstNameBox" required/></td>
            </tr>
            <tr>
                <td><label for="LastNameLabel" class="tableproperties">Last Name</label></td>
                <td><input type="text" name="lastName" placeholder="Enter Last Name" id="LastNameBox" class="signupTextBoxStyle" required></td>
            </tr>
        <tr>
            <td><label for="eMailLabel" class="tableproperties">Email</label></td>
            <td><input type="email" name="email" placeholder="Enter Email" id="eMailBox" class="signupTextBoxStyle" required></td>
            <td id="emailStatus"></td>
        </tr>

        <tr>
            <td><label for="passwordLabel" class="tableproperties">Password</label></td>
            <td><input type="password" name="password" placeholder="Enter Password" id="passwordTextbox" maxlength="24" class="signupTextBoxStyle" required></td>
            <td><i class="fa fa-info-circle infoIcon" title="Password must contain minimum 3 upper case, 2 lower case and 2 special chars"></i></td>
            <td><progress value="0" max="100" class="progressBar" id="progressStatus"></progress></td>
            <td id="passwordStrength"></td>
        </tr>

        <tr>
            <td><label for="confirmPasswordLabel" class="tableproperties">Confirm Password</label></td>
            <td><input type="password" name="confirmpassword" placeholder="Must be same as password" maxlength="24" id="confirmPasswordBox" class="signupTextBoxStyle" required></td>
            <td id="passwordMismatch"></td>
        </tr>

        <tr>
            <td><label for="dobLabel" class="tableproperties">D.O.B</label></td>
            <td><input type="date" name="dob" placeholder="Enter D.O.B" id="dobBox" class="signupTextBoxStyle" required></td>
        </tr>

        <tr>
            <td><label for="dobTimeLabel" class="tableproperties">D.O.B with time</label></td>
            <td><input type="datetime" name="dobTime" placeholder="Enter D.O.B with time" id="dobTimeBox" class="signupTextBoxStyle" required></td>
        </tr>

        <tr>
            <td><label for="localDOBLabel" class="tableproperties">Local D.O.B</label></td>
            <td><input type="datetime-local" name="localdob" placeholder="Enter Local D.O.B" id="localDobBox" class="signupTextBoxStyle" required></td>
        </tr>

        <tr>
            <td><label for="ssnLabel" class="tableproperties">SSN</label></td>
            <td><input type="text" name="ssn" placeholder="000-00-0000" id="ssnBox" class="signupTextBoxStyle" required pattern="^(\d{3}-\d{2}-\d{4})$"></td>
        </tr>

        <tr>
            <td><label for="usPhoneNumber" class="tableproperties" >US Phone Number</label></td>
            <td><input type="text" name="phone" placeholder="000-000-0000" id="usNumberBox" class="signupTextBoxStyle" required></td>
            <td id="phoneStatus"></td>
        </tr>

        <tr>
            <td><label for="creditLabel" class="tableproperties" id="CreditText">Credit Card Number</label></td>
            <td><input type="text" name="creditCardNumber" placeholder="Enter Credit Card Number" id="creditBox" class="signupTextBoxStyle" required pattern="^[0-9]{12}(?:[0-9]{4})?$"></td>
        </tr>


        <tr>
            <td colspan='2'>
                <input type="submit" class="btn btn-primary btn-lg btn-block signupbuttonStyle" id="sub" />
                <button type="button" class="btn btn-danger btn-lg btn-block signupbuttonStyle" onclick="location.href = 'index.html';">Cancel</button>
            </td>
        </tr>
        </table>
    </form>

PHP （只是为了测试数据如果我手动输入数据就会保存到mySQL中）

$json_obj = '{
      "jsonFirstName": "Kishan",
      "jsonLastName": "Kishan",
      "jsonEmail": "Kishan",
      "jsonPassword": "Kishan",
      "jsonDob": "Kishan",
      "jsonDobTime": "Kishan",
      "jsonLocaldob": "Kishan",
      "jsonSsn": "Kishan",
      "jsonPhonenumber": "Kishan",
      "jsonCreditcardnumber": "Kishan"
 }';

PHP （如果我想从表单中获取值，则会出错）

$json_obj = '{
      "jsonFirstName": (string) $_POST['firstName'],
      "jsonLastName": (string) $_POST['lastName'],
      "jsonEmail": (string) $_POST['email'],
      "jsonPassword": (string) $_POST['password'],
      "jsonDob": (string) $_POST['dob'],
      "jsonDobTime": (string) $_POST['dobTime'],
      "jsonLocaldob": (string) $_POST['localdob'],
      "jsonSsn": (string) $_POST['ssn'],
      "jsonPhonenumber": (string) $_POST['phone'],
      "jsonCreditcardnumber": (string) $_POST['creditCardNumber']
 }';

错误说明
 解析错误：语法错误，第19行/Applications/XAMPP/xamppfiles/htdocs/xampp/297test/userInfo.php中的意外'firstName'（T_STRING）

PHP代码的REST
$ result = json_decode（$ json_obj）; 

$firstname = $result->jsonFirstName;
$lastname = $result->jsonLastName;
$email = $result->jsonEmail;
$password = $result->jsonPassword;
$dob = $result->jsonDob;
$dobTime = $result->jsonDobTime;
$localdob = $result->jsonLocaldob;
$ssn = $result->jsonSsn;
$phonenumber = $result->jsonPhonenumber;
$creditcardnumber = $result->jsonCreditcardnumber;


if(mysql_query("INSERT INTO user VALUES('$firstname', '$lastname', '$email', '$password', '$dob', '$dobTime', '$localdob', '$ssn','$phonenumber','$creditcardnumber')")){
    echo "Successfully Inserted";
}

else
    echo "Fail to Insert";

4 个答案:

答案 0 :(得分：1)

由于引号，换行符等原因，直接通过连接创建JSON-Strings很难。

相反，创建一个值数组并使用json_encode将其编码为JSON字符串：

$values = array(
  "jsonFirstName" =>        $_POST['firstName'],
  "jsonLastName" =>         $_POST['lastName'],
  "jsonEmail" =>            $_POST['email'],
  "jsonPassword" =>         $_POST['password'],
  "jsonDob" =>              $_POST['dob'],
  "jsonDobTime" =>          $_POST['dobTime'],
  "jsonLocaldob" =>         $_POST['localdob'],
  "jsonSsn" =>              $_POST['ssn'],
  "jsonPhonenumber" =>      $_POST['phone'],
  "jsonCreditcardnumber" => $_POST['creditCardNumber']
);

$json_obj = json_encode($values);

或者你可以这样做：

$json_obj = json_encode($_POST);

然后，您将获得每个索引为$_POST的JSON对象。唯一的区别是，您不能像在示例中那样重命名字段。

答案 1 :(得分：0)

不确定为什么这样做，但是解析错误是因为你使用单引号打开字符串然后使用它来选择数组索引。

更改$_POST['firstName']以使用双引号，例如$_POST["firstName"]

答案 2 :(得分：0)

编辑： 试着这样做。

$array = (
      "jsonFirstName" =>  $_POST['firstName'],
      "jsonLastName" =>  $_POST['lastName'],
      "jsonEmail" => $_POST['email'],
      "jsonPassword" => $_POST['password'],
      "jsonDob" => $_POST['dob'],
      "jsonDobTime" => $_POST['dobTime'],
      "jsonLocaldob" => $_POST['localdob'],
      "jsonSsn" => $_POST['ssn'],
      "jsonPhonenumber" => $_POST['phone'],
      "jsonCreditcardnumber" => $_POST['creditCardNumber']
 );

$json = json_encode($array);

答案 3 :(得分：0)

您收到错误的原因是您以错误的方式从JSON变量中创建POST对象。您的陈述中确实存在引号问题。它应该是这样的：

  $json_obj = '{
  "jsonFirstName": "'.(string) $_POST['firstName'].'",
  "jsonLastName": "'.(string) $_POST['lastName'].'",
  "jsonEmail": "'.(string) $_POST['email'].'",
  "jsonPassword": "'.(string) $_POST['password'].'",
  "jsonDob": "'.(string) $_POST['dob'].'",
  "jsonDobTime": "'.(string) $_POST['dobTime'].'",
  "jsonLocaldob": "'.(string) $_POST['localdob'].'",
  "jsonSsn": "'.(string) $_POST['ssn'].'",
  "jsonPhonenumber": "'.(string) $_POST['phone'].'",
  "jsonCreditcardnumber": "'.(string) $_POST['creditCardNumber'].'"
  }';

如果我在构造JSON对象时只包含没有双引号的变量，我会得到类似的结果：

  $name = "Foo";
  $json_obj = '{"firstName" :'.$name.'}'; 

 //gets expanded to $json_obj = {"firstname" : Foo}
 //Whereas it should be $json_obj = {"firstname" :"Foo"}

由于缺少双引号，您无法在第一时间创建JSON对象，因此当您尝试对其进行解码并进一步访问属性时，它会出现错误提示：

 Trying to get property of non-object in /Applications/XAMPP/xamppfiles/htdocs/xampp/297test/userInfo.php on line 35

尝试我提供的代码段，希望能帮到你。

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