如何从CodeIgniter中插入包含连字符/破折号的电子邮件 - How to insert emails containing hyphen/dash in mysql from CodeIgniter?

我正面临一个问题插入电子邮件与连字符/破折号到mysql。使用的框架是CodeIgniter。项目托管在GoDaddy中(如果这有帮助的话)。有效的电子邮件是:

name@domain.com, name@test.domain.com, first.last@domain.com，first.last@test.domain.com, name.123@domain.com, first_last@domain.com, first_last@test.domain.com

不工作的电子邮件是，但他们工作良好的localhost:

first-last@domain.com, first-last@test.domain.com

下面是插入电子邮件的表单:

<form method="post" action="<?php echo base_url('index.php?/codes');?>">
      <div class="form-group">
          <label>Email</label>
          <input type="text" name="email" id="email" placeholder="Email" class="form-control">
      </div>
      <div class="form-group">
          <label>Waiver Code</label>
          <input type="text" name="code" id="code" placeholder="Code" class="form-control">
      </div>
      <button class="btn btn-sm btn-success" onclick="autogen()" name="saveCode">Generate</button>
</form>

从'javascript'调用autogen()函数:

function autogen() {
    var randomstring = Math.random().toString(36).slice(-6);
    var date = new Date().getFullYear();
    randomstring = date+randomstring;
    randomstring = randomstring.toUpperCase();
    var email = $('#email').val();
    var code = $('#code');
    if (!email) {
        alert("Email is required!");
        $('#email').focus();
    } else {
        code.val(randomstring);
        alert("Email: "+email+"'nCode: "+randomstring); 
        
        $.ajax({
          url: "<?php echo base_url('index.php?/genCode/"+email+"/"+randomstring+"');?>",
          data: ({'email': email, 'code': randomstring}),               
          type: "post",
          success: function(response, textStatus, jqXHR){
            location.reload();
            alert('Code added');
          },
          error: function(jqXHR, textStatus, errorThrown){
            console.log("The following error occured: "+
                        textStatus, errorThrown);
          }
        });
    }
}

最后是CodeIgniter的插入脚本

class GenCode extends CI_Controller {
public function index($email="", $code="")
{
    //$data = array('email' => $this->db->escape($email), 'code' => $code, 'user' => $this->session->userdata('username'));
    //$query = $this->db->insert('codes', $data);
$query = $this->db->query("insert ignore into codes(email, code, user) values('".$this->db->escape($email)."', '".$code."', '".$this->session->userdata('username')."');");
    if ($query == TRUE)
        return TRUE;
    else
        return FALSE;
}
}

我尝试过但没有运气的事情:

$ this→db→逃脱(电子邮件)
mysql_real_escape_string(电子邮件)

我不知道我哪里错了。还是跟GoDaddy有关?

我假设您正在使用PDO。这个例子是针对PDO的:

$stmt=$this->db->prepare("insert ignore into codes(email, code, user) ".
   " values(:email, :code, :user)");
$stmt->bindParam(":email",$email);
$stmt->bindParam(":code",$code);
$stmt->bindParam(":user",$user);
$stmt->execute();

我不得不重新工作你的代码相当多，但我运行这个，它工作得很好。首先是表格。您不需要提交表单和执行AJAX，只执行其中一个即可。因为简单总是更好，所以我选择提交表单。下面是为了测试而更改了控制器和方法的表单。控制器名称为email_test，方法为insert email。还要注意(非常重要)onclick中的Javascript调用有一个附加的"return"。这样，如果email字段中没有值，表单就不会提交。

<form method="post" action="email_test/insert_email">
    <div class="form-group">
        <label>Email</label>
        <input type="text" name="email" id="email" placeholder="Email" class="form-control">
    </div>
    <div class="form-group">
        <label>Waiver Code</label>
        <input type="text" name="code" id="code" placeholder="Code" class="form-control">
    </div>
    <button class="btn btn-sm btn-success" onclick="return autogen();" name="saveCode">Generate</button>
</form>

接下来是Javascript。我将其简化为检查电子邮件和生成代码。还要注意注释。

function autogen() {
    var randomstring = Math.random().toString(36).slice(-6);
    var date = new Date().getFullYear();
    randomstring = date + randomstring;
    randomstring = randomstring.toUpperCase();
    var email = $('#email').val();
        /*
         * This checks for a value, but it does not mean its an email.
         * You need a valid email check here too.
         */
    if (!email) {
        alert("Email is required!");
        $('#email').focus();
        /*
         * form will not submit
         */
        return false;
    } else {
        $('#code').val(randomstring);
        console.log("Email: " + email + "'nCode: " + randomstring);
        /*
         * form will submit
         */
        return true;
    }
}

最后是控制器和模型。为简洁起见，我跳过了模型，但您不应该这样做。查看下面的评论

class Email_test extends CI_Controller {
    public function __construct() {
        parent::__construct();
    }
    public function index() {
    }
    public function insert_email() {
        /*
         * TODO here...
         * check for ajax
         * check for being sent from yourself
         * clean your input
         * return error and handle if bad input
         */
        // load your model here
        // call your model here which has the following code
        //all this code should be in a model, put here for an example.
        $query = $this->db->query("insert into codes(email, code, user) values(" . $this->db->escape($email) . ", '$code.', '" . $this->session->userdata('username') . "');");
        if ($this->db->affected_rows() > 0) {
            return TRUE;
        } else {
            return FALSE;
        }
        //handle the return from the model here
    }
}