从HTML表单插入输入数据到JSON对象，然后将其存储在MYSQL中

我想从HTML中的用户注册表单获取数据，然后将数据推送到JSON，然后获取JSON并存储到MySQL中。请帮帮我。从HTML表单插入输入数据到JSON对象，然后将其存储在MYSQL中

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（只是为了测试数据得到保存到MySQL，如果我手动输入数据）

$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（错误从表格中获取数值）

$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，意想不到在/Applications/XAMPP/xamppfiles/htdocs/xampp/297test/userInfo.php '姓'（T_STRING）上线的PHP 19

REST代码
$ 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";

来源

2014-10-07 Kishan Chaitanya

你为什么要在json中转换它？ – 2014-10-07 06:10:42

我不明白你为什么需要将POST变量放入json字符串中，然后解码，然后再次使用它。它没有意义 – Ghost 2014-10-07 06:10:58

$ json_obj =“{'jsonFirstName'：”。 mysqli_real_escape_string（$ _ POST ['firstName']。“）} - 或者更好的是，使用准备好的语句... – FAS 2014-10-07 06:13:29

直接通过串联创建JSON字符串很难，因为引号，换行符等等。

相反，创造价值的数组，编码成JSON字符串json_encode：

$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);

另外，你可以这样做：

$json_obj = json_encode($_POST);

然后，您将得到每一个JSON对象索引$_POST。唯一的区别是，你不能像在你的例子中那样重命名你的字段。

来源

2014-10-07 10:51:03 FAS

谢谢...迄今它的gng on gud ....当我回显$ json_obj ..我得到下面的结果（{“ jsonFirstName “：” userfirst”， “jsonLastName”： “userlast”， “jsonEmail”： “[email protected]”， “jsonPassword”： “QWERqwer @＃$”， “jsonDob！”： “2014年1月1日” “jsonDobTime”： “dobtime”， “jsonLocaldob”： “2014-01-01T01：00”， “jsonSsn”： “123-12-1234”， “jsonPhonenumber”： “123-123-1234”， “jsonCreditcardnumber” ：“123412341234”}）...你能请指导我如何在mySQl服务器中逐一存储这些值...这将实际上解决我的问题.. – 2014-10-08 01:50:26

它的工作感谢很多fabian ... – 2014-10-08 03:04:47

不知道你为什么这样做，但是解析错误是因为你使用的是单引号打开字符串，然后又使用该挑数组索引。

变化$_POST['firstName']用双引号如$_POST["firstName"]

来源

2014-10-07 06:12:30 Lock

它stil不wrk ...这个错误进来注意：试图获取非对象的属性在/ Applications/XAMPP/xamppfiles /htdocs/xampp/297test/userInfo.php on line 35“$ firstname = $ result-> jsonFirstName;” - >这行指出 – 2014-10-07 06:18:03

编辑：尝试这样做。

$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);

来源

2014-10-07 06:19:29

仍然没有运气..现在的错误是“解析错误：语法错误，第18行中的/Applications/XAMPP/xamppfiles/htdocs/xampp/297test/userInfo.php中出现意外的'{' – 2014-10-07 06:23:50

看看我编辑的问题 – 2014-10-07 06:32:33

解析错误：语法错误，意外'=>'（T_DOUBLE_ARROW）in/Applications/XAMPP/xamppfiles/htdocs/xampp/297test/userInfo.php ..我得到这个新的错误:( – 2014-10-07 06:43:14

，因为你是在一个错误的方式创建JSON对象了POST变量你得到一个错误。在你的陈述中引用确实存在问题。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'].'" 
    }';

如果我只包括没有双引号的变量而构建JSON对象，我得到这样的：

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

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

正是因为缺少双引号的，您无法首先创建JSON对象，因此当您尝试对其进行解码并进一步访问某个属性时，它会给出错误提示：

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

试试我给的代码片段，它应该有希望帮助你。

来源

2014-10-07 08:57:27

从HTML表单插入输入数据到JSON对象，然后将其存储在MYSQL中

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