PHP isset函数未正确检查

问题描述 投票:-2回答:3

单击按钮时。我想使用register_button从表单中获取值。但是,它不会进入if子句。

我无法在if语句中检查表单中输入的任何内容。有没有其他选项我可以从表单中检索数据或者我是否犯了一些错误?

var_dump($_POST)回归

array(6) { 
    ["reg_fname"]=> string(1) "a" 
    ["reg_lname"]=> string(3) "asd" 
    ["reg_email"]=> string(5) "ad@ad" 
    ["reg_pass"]=> string(4) "dasd" 
    ["reg_pass2"]=> string(5) "sadsa" 
    ["register_button"]=> string(8) "Register" 

} 

Invalid format 

码:

$fname = "";
$lname = "";
$em = "";
$em2 = "";
$password = "";
$password2 = "";
$date = "";
$error_array = "";

if (isset($_POST['register_button'])) {

    $fname = strip_tags($_POST['reg_fname']);
    $fname = str_replace(' ', ' ', $fname);
    $lname = $_POST['reg_lname'];
    $lname = str_replace(' ', ' ', $lname);

    $em = $_POST['reg_email'];
    $em = str_replace(' ', ' ', $em);

    $password = strip_tags($_POST['reg_pass']);
    $password2 = strip_tags($_POST['reg_pass2']);

    $date = date("Y-m-d");

    if(filter_var($em, FILTER_VALIDATE_EMAIL)){
        $em = filter_var($em, FILTER_VALIDATE_EMAIL);

        $e_check = mysqli_query($con, "SELECT email FROM users WHERE email='$em");

        $num_rows = mysqli_num_rows($e_check);

        if(num_rows > 0){
            echo "Email already in use";
        }
    }else {
        echo "Invalid format";
    }

    if(strlen(fname)>25 || strlen(fname)<2 ){
        echo "Your fi";
    }
}

<form method="post" action="index.php">

<input type = "text" name="reg_fname" placeholder="First Name"       required>
<br>
<input type = "text" name="reg_lname" placeholder="Last Name" required>
<br>
<input type = "email" name="reg_email" placeholder="Email" required>
<br>
<input type = "password" name="reg_pass" placeholder="Password" required>
<br>
<input type = "password" name="reg_pass2" placeholder="Confirm Password" required>
<br>
<input type = "submit" name="register_button" value="Register">

</form>
php html post isset
3个回答
0
投票

它正在进入第一个if声明。你可以告诉,因为当你var_dump($_POST)你在转储输出后得到文本“无效的格式”。这意味着它进入你的else声明进一步向下页面,你有echo "Invalid format";。所以它实际上是你的filter_var if声明失败了。


0
投票

为什么使用“register_button”作为支票?为什么不是你的一个领域是必要的?


0
投票

任何人回答后请不要编辑您的问题您不检查查询是否成功。你使用if过滤变量最好是直接使你忘了在字符串后面的查询中添加逗号。

请尝试以下

<?php
$fname = "";
$lname = "";
$em = "";
$em2 = "";
$password = "";
$password2 = "";
$date = "";
$error_array = "";
if (isset($_POST['register_button'])) {
    $fname = strip_tags($_POST['reg_fname']);
    $fname = str_replace(' ', ' ', $fname);
    $lname = $_POST['reg_lname'];
    $lname = str_replace(' ', ' ', $lname);

    $em = $_POST['reg_email'];
    $em = str_replace(' ', ' ', $em);

    $password = strip_tags($_POST['reg_pass']);
    $password2 = strip_tags($_POST['reg_pass2']);

    $date = date("Y-m-d");
    // No need to use if
    $em = filter_var($em, FILTER_VALIDATE_EMAIL);
    /// $con = mysqli_connect("localhost", "user", "pass", "dbname");
    /// using if to be sure query success
    if ($e_check= mysqli_query($con, "SELECT email FROM users WHERE email='$em'")) {

        $num_rows= mysqli_num_rows($e_check);
        if ($num_rows> 0) {
            echo "Email already in use";
        } else {
            /// here your insert query

        }
    }
    else
    {
        echo "No query ";
    } 
    if(strlen($fname) >25 || strlen($fname)< 2 )
    { 
        echo "Your first name must be at least 2 char and maximum 25 char "; 
    }
}
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