dongwenhui8900 2015-09-05 07:18 采纳率: 0%
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使用PHP创建图像验证码

Hi i want to create a image captcha with php, i have the next code for my form.

<form action="" method="POST">
    <div class="label_form">Usuario:</div> <input type="text" name="user"/><br>
    <div class="label_form">Contraseña:</div> <input type="password" name="pass"/><br>  
    <img alt="Numeros aleatorios" src="layouts/captcha.php" />  
    <input class="label_form" type="text" name="num"/><br>
    <input type="submit" value="ENTRAR" name="submit"/>
</form> 

This is the code for validation before to send the form:

if (isset($_POST["submit"])) {
  if ($_SESSION['img_number'] != $_POST['num']) {
    echo "<div class='msg_error'>Los caracteres no se corresponden.</div>";
  } else {
    /*DO STUFF*/
  }
}

And in other file with the name captcha.php i have the code phpfor generate the image:

header("Content-type: image/png");
$string = "abcdefghijklmnopqrstuvwxyz0123456789";
for ($i = 0; $i < 5; $i++) {
  $pos = rand(0, 36);
  $str. = $string {
    $pos
  };
}
$img_handle = ImageCreate(60, 22) or die("Es imposible crear la imagen");
$back_color = ImageColorAllocate($img_handle, 102, 102, 153);
$txt_color = ImageColorAllocate($img_handle, 255, 255, 255);
ImageString($img_handle, 31, 5, 0, $str, $txt_color);
Imagepng($img_handle);
session_start();
$_SESSION['img_number'] = $str;

This give a image broken showing the alt from img "Numeros aleatorios", so that tell me the file.php is calling fine the img but the code for generated is not working, any help is gratefull :D thanks.

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3条回答 默认 最新

  • doutandusegang2961 2015-09-05 14:33
    关注

    Your code is good the only thing you missed is declaring $str = "" at the top of your catpcha.php file

    Here is a working capture of your code:

    enter image description here

    Update

    Your $string length is 36, so you should generate a random position between 0 and 35, like this:

    $pos = rand(0,35);
    

    If not, you sometimes get a broken image and a PHP Notice.

    Update N°2

    The reason why you get a broken image without declaring $str = "" is because this line

    $str .= $string{$pos};
    

    Throws a PHP Notice and corrupts the image. (Of course only if you display_errors , that's why it worked perfectly for @Danny's server)

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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