PHP 内爆没有通过

Question

我收到一个错误

Warning: implode():Invalid arguments passed

我不明白为什么会这样...

这是我的代码。

if(isset($_POST['consultationbutton'])){        
    $image = addslashes(file_get_contents($_FILES['selectedfile']['tmp_name'])); //SQL Injection defence!
    $checkedcondition = implode(",",$_POST['skincondition']);
    $checkedproduct = implode(",",$_POST['skincareinuse']);
    $consultquery="INSERT INTO counsel(nric,dateconsulted,editableface,skincarecurrentlyinuse,skincondition) VALUES('132','$_POST[storedate]','{$image}','$checkedproduct','$checkedcondition')";
    mysqli_query($dbconn,$consultquery);
}

$_POST['skincondition'] 和 $_POST['skincareinuse'] 是我的复选框的名称。

Answer 1

HTML 部分

<form action="" method="post">
<input type="checkbox" name="skincondition[]" value="skincondition_1" />skincondition_1

<input type="checkbox" name="skincondition[]" value="skincondition_2" />skincondition_2

<input type="checkbox" name="skincondition[]" value="skincondition_3" />skincondition_3

<input type="checkbox" name="skincondition[]" value="skincondition_4" />skincondition_4

<input type="checkbox" name="skincondition[]" value="skincondition_5" />skincondition_5

<input type="submit" name="consultationbutton">
</form>

PHP代码

if(isset($_POST['consultationbutton'])){        
$checkedcondition = implode(",",$_POST['skincondition']);
echo $checkedcondition;
}

对我来说工作正常，请分享您的 HTML 代码以便我更好地告诉您。

PHP 内爆没有通过

PHP implode not getting passed

php

implode