表单是一个简单的验证码，但它无法验证我的if语句中的第一个条件

Question

这是第一个将被保存为image.php的php脚本。 这是代码：

<?php
$img = imagecreatetruecolor(200,80);

$white = imagecolorallocate($img, 255, 255, 255);
$black = imagecolorallocate($img, 0, 0, 0);
$red = imagecolorallocate($img, 255, 0, 0);
$grey = imagecolorallocate($img,150, 150, 150);
$pink = imagecolorallocate($img, 200, 0, 150);

function randomString($length){
    $chars = "abcdefghijkmnopqrstuvwxyz023456789";
    srand((double)microtime()*1000000);
    $str = "";
    $i = 0;

        while($i <= $length){
            $num = rand() % 33;
            $tmp = substr($chars, $num, 1);
            $str = $str . $tmp;
            $i++;
        }
    return $str;
}

for($i=1;$i<=rand(1,5);$i++){
    $color = (rand(1,2) == 1) ? $pink : $red;
    imageline($img,rand(5,70),rand(5,20), rand(5,70)+5,rand(5,20)+5, $color);
}

imagefill($img, 0, 0, $white);

$string = randomString(rand(2,5));
$_SESSION['string'] = $string;

imagettftext($img, 11, 0, 10, 20, $black, "calibri.ttf", $string);

header("Content-type: image/png");
imagepng($img);
imagedestroy($img);
?>

在上面的代码中一切正常。...现在我创建一个新的php脚本并将其命名为captcha.php，然后将此代码插入脚本中：

<?php
ob_start();
session_start();
if(!isset($_POST['submit'])){
    echo "<form method=\"post\" action=\"captcha.php\">\n";
    echo "<table border=\"0\" cellspacing=\"3\" cellpadding=\"3\">\n";
    echo "<tr><td>Type The Letters You See Below Into the Box</td></tr>\n";
    echo "<tr><td align=\"right\"><img src=\"image.php\"></td></tr>\n";
    echo "<tr><td align=\"right\"><input type=\"text\" name=\"image\"></td></tr>\n";
    echo "<tr><td align=\"right\"><input type=\"submit\" name=\"submit\" value=\"Check CAPTCHA\"></td></tr>\n";
    echo "</table></form>\n";
}else {
    $image = $_POST['image'];
    if ($image == $_SESSION['string'])
    {
        echo "<b>Great success!</b>\n";
    }
    else 
    {
        echo "<em>Failure!</em>\n";
    }
}
ob_end_flush();
?> 

过去我遇到的困难是： if ($image == $_SESSION['string'])它无法正确验证。

Answer 1

您还必须在image.php中开始会话。