验证码怎么验证？-tutorial php-php.cn

验证码怎么验证？

WBOY

Lepaskan： 2016-07-06 13:53:52

asal

1474 orang telah melayarinya

弄了一个简答的验证码最开始只有一个验证码的输入框和验证码图片提交后可以打印出来你输入的验证码是正确的还是错误的后来我又加了一个文本框加了几句sql语句想做个简单的验证提交可是不知道该怎么验证在实际中肯定不只是提示错误或者正确吧如果错误应该怎么阻止表单提交呢？还有一个需要先解决的问题是这个验证码本身不需要插入到数据库中但是验证码的表单中有name啊如果sql语句不处理就点击提交就会报错这个应该怎么办呢？ Invalid parameter number: number of bound variables does not match number of tokens in D:wampwwwyzm1.php on line 28

<code>


    <meta charset="utf-8">
    <title></title>


<form method="post">
    <input type="text" name="num"><br>
    <input type="text" style="display: inline-block;width: 50px;" name="yzm">
    <img  id="checkpic" onclick="javascript:this.src='yzm2.php?tm='+Math.random();" src="yzm2.php" alt="验证码怎么验证？" >
    <button type="submit">提交</button>
</form>
<?php if(trim(@$_POST['num'])){
        session_start();
        if(isset($_POST['yzm'])){
            if($_SESSION["str"]==$_POST['yzm']){
              echo "right";
            }else{
              echo "wrong";
          }
        }
        
        $pdo=new PDO("mysql:host=localhost;dbname=t1","root","");
        $stmt=$pdo->prepare("insert into test(num)values(:num)");
        $stmt->execute($_POST);
    }
    
?>


</code>

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回复内容：

<code>


    <meta charset="utf-8">
    <title></title>


<form method="post">
    <input type="text" name="num"><br>
    <input type="text" style="display: inline-block;width: 50px;" name="yzm">
    <img  id="checkpic" onclick="javascript:this.src='yzm2.php?tm='+Math.random();" src="yzm2.php" alt="验证码怎么验证？" >
    <button type="submit">提交</button>
</form>
<?php if(trim(@$_POST['num'])){
        session_start();
        if(isset($_POST['yzm'])){
            if($_SESSION["str"]==$_POST['yzm']){
              echo "right";
            }else{
              echo "wrong";
          }
        }
        
        $pdo=new PDO("mysql:host=localhost;dbname=t1","root","");
        $stmt=$pdo->prepare("insert into test(num)values(:num)");
        $stmt->execute($_POST);
    }
    
?>


</code>

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<code><?php if(trim(@$_POST['num'])){
        session_start();
        if(isset($_POST['yzm'])){
            if($_SESSION["str"]==$_POST['yzm']){
                echo "right";
                $pdo=new PDO("mysql:host=localhost;dbname=t1","root","");
                $stmt=$pdo->prepare("insert into test(num)values(:num)");
                $stmt->execute(array(':num'=>$_POST['num']);
            }else{
                echo "wrong";
          }
        }
    }
?>
</code>

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这样写即可