我有生成安全密码的功能。点击一个按钮，我进行了一个生成密码的ajax调用。当我尝试返回值，字符串或其他东西时，ajax不会通过访问，但日期会在数据库中更新...我无法找到无法正常工作的地方。
<input type="button" value="generate pass" class="btn btn-info mr50 auth-key-button" onclick="serialkey(<?=$_GET['id']?>)">
Ajax: <script>
document.querySelector(".auth-key-button").addEventListener("click", function(){
document.querySelector(".auth-key-inputbox").style.display = "block";
});
function serialkey(id) {
$.ajax({
type: "POST",
url: 'generator.php',
dataType: 'text/html',
data:{
id: id,
action:'call_this'
},
success:function(data) {
alert(2);
$( "#result" ).show();
$( "#divKey" ).hide();
}
});
}
File generator.php where it generates the password:
$user = dbUse( "SELECT * FROM users WHERE id = ".$_POST['id']." ORDER BY id ASC;" );
if($_POST['action'] == 'call_this') {
function incrementalLetter($len = 1){
$charset = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
$base = strlen($charset);
$result = '';
$now = explode(' ', microtime())[1];
while ($now >= $base){
$i = $now % $base;
$result = $charset[$i] . $result;
$now /= $base;
}
return substr($result, -1);
}
function random($length, $chars = '')
{
if (!$chars) {
$chars = implode(range('a','f'));
$chars .= implode(range('0','9'));
$time = -microtime(true);
$hash = 0;
for ($i=0; $i < rand(1000,4000); ++$i) {
$hash ^= sha1(substr(str_shuffle("0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"), 0, rand(1,10)));
}
$chars .= $time + microtime(true);
}
$shuffled = str_shuffle($chars);
$shuffled = str_replace('.', incrementalLetter(), $shuffled);
return substr($shuffled, 0, $length);
}
$key = random(6).'-'.random(6).'-'.random(6).'-'.random(6).'-'.random(6).'-'.random(6);
$unique = dbUse( "SELECT auth_key, COUNT(*) c FROM users GROUP BY auth_key HAVING c>1;" );
if(count($unique)>0){
random(6).'-'.random(6).'-'.random(6).'-'.random(6).'-'.random(6).'-'.random(6);
}else{
dbUse( "UPDATE users SET auth_key = '".$key."' WHERE id= ".$_POST['id'].";" );
}
return $key;
}
Console.WriteLine("A");
strings.ForEach( s => { AsyncMethod(s).Wait(); });
Console.WriteLine("E");
答案 0 :(得分：1)
有效...这就是我找到解决方案的方式，如果其他人需要它
$data = "";
$data['key'] = $key;
echo json_encode($data);
我从ajax中删除dataType：'text / html'，我使用var data = JSON.parse（data）; 那是......