代码:
<?php
if($this->input->post('send'))
{
$email = $this->input->post('email');
$this->db->select('password,firstname');
$this->db->from('user');
$where = "email = '$email'";
$this->db->where($where);
$query = $this->db->get();
if ($query->num_rows() > 1)
{
$result = $query->result_array();
foreach ($result as $row)
{
$password = $row['password'];
$firstname = $row['firstname'];
}
if($result)
{
$toEmail = $this->input->post('email');
$fromEmail = "info@xyz.com";
$this->load->library('email');
$this->email->from($fromEmail, 'test');
$this->email->to($toEmail);
$this->email->subject('user Login Password');
$body="Hi ".$firstname."Your Login Password is ".$password."";
$this->email->message($body);
$this->email->send();
if (!$this->email->send())
{
print_r($this->email->print_debugger(), true);
}
else
{
echo '<p style="color: red;font-weight: bold;">Error!.</p>';
}
echo '<p style="color: #41b212;font-weight: bold;">Kindly check your email to get your password.</p>';
}
}
else
{
echo '<p style="color: red;font-weight: bold;">Your email id does not exist.</p>';
}
}
?>
<form method="post">
<div class="form-group">
<input type="text" name="email" id="email" placeholder="Enter Your Email" class="text-line"/>
</div>
<div class="form-group">
<input type="submit" name="send" id="send" value="Send" class="btn btn-warning"/>
</div>
</form>
在此代码中,我创建了一个重置密码表单。在这里,我有一个名为user的表。现在,当我点击发送按钮时,它始终显示您的电子邮件ID不存在。但是我正在创建如果用户电子邮件ID已经存在,它将向用户表中具有电子邮件ID的用户发送邮件。那么,我该怎么办呢?请帮帮我。
谢谢
答案 0 :(得分:3)
您正在检查的情况有问题 变化
if ($query->num_rows() > 1)
到
if ($query->num_rows() > 0)