【发布时间】：2014-12-01 11:58:18
【问题描述】：
我正在尝试从 HTML 中的用户注册表单中获取数据，然后将数据推送到 JSON，然后获取 JSON 并存储到 MySQL。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（只是为了测试如果我手动输入数据，数据会保存到 mySQL）
$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";
【问题讨论】：
-
为什么要转成json？
-
我不明白您为什么需要将 POST 变量转换为 json 字符串，然后对其进行解码，然后再次使用它。这没有意义
-
$json_obj = "{'jsonFirstName': " 。 mysqli_real_escape_string($_POST['firstName'] . ")} - 或者更好的是，使用准备好的语句...
-
我只是在我的大学做作业..这些是要求...我需要在 json 中获取表单输入，然后将其解析到 SQL server...
-
@Fabian 它仍然没有工作。感谢您的输入:(