與分佈數據

Question

存儲的個人資料相片所以，我目前所面對的這個代碼，但它只能存儲PIC的文件夾中，並沒有超越這一步。 表單中輸入的數據不會被存儲在數據庫.... :(

<?php 
$sub=0; 
ini_set("display_errors", 0); 
if(isset($_REQUEST['submited'])) { 

// your save code goes here 

$allowedExts = array("jpg", "jpeg", "gif", "png"); 
$extension = end(explode(".", $_FILES["file"]["name"])); 
if ((($_FILES["file"]["type"] == "image/gif") 
|| ($_FILES["file"]["type"] == "image/jpeg") 
|| ($_FILES["file"]["type"] == "image/png") 
|| ($_FILES["file"]["type"] == "image/pjpeg")) 
&& ($_FILES["file"]["size"] < 2097152) 
&& in_array($extension, $allowedExts)) 
{ 
if ($_FILES["file"]["error"] > 0) 
{ 
echo "Return Code: " . $_FILES["file"]["error"] . "<br>"; 
} 
else 
{ 
echo ""; 
if (file_exists("media/" . $_FILES["file"]["name"])) 
{ 
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload already exists.</b></font>"; 
} 

else 
{ 
move_uploaded_file($_FILES["file"]["tmp_name"], 
"media/" . $_FILES["file"]["name"]); 
$sub= 1; 
$mysqli = new mysqli("localhost", "root", "", "cms"); 

// TODO - Check that connection was successful. 

$name = $_POST["name"]; 
$surname = $_POST["surname"]; 
$about = $_POST["about"]; 
$visible = $_POST["visible"]; 
$admin = $_POST["admin"]; 

$file_path = $_FILES['file']['name']; 
$type = $_FILES['file']['type']; 
$size = $_FILES['file']['size']; 


$stmt = $mysqli->prepare("INSERT INTO schauspieler (
name, surname, about, visible, admin, date, file_path, photo_type, photo_size 
) VALUES (
'$name', '$surname', '$about', $visible, $admin, '$file_path', '$type', '$size', CURRENT_TIMESTAMP, 
)"); 

// TODO check that $stmt creation succeeded 

// "s" means the database expects a string 
$stmt->bind_param("s", $name, $surname, $about, $visible, $admin, $file_path, $type, $size); 

$stmt->execute(); 

$stmt->close(); 

$mysqli->close(); 



echo "<font size='7' color='red'><b> Success! Your photo has been uploaded.</b></font>"; 
} 

} 
} 
else 
{ 
echo "<font size='4' color='red'><b>We are sorry, the file you trying to upload is not an image or it exceeds 2MB in size.</b></font><br><font color='blue'><i>Only images under size of 2MB are allowed</i></font>."; 
} 
} 
?> 

誰能告訴我，我錯過了什麼？

感謝名單傢伙

乾杯 克里斯

Answer 1

看來你有你準備好的聲明中的一個問題。

查詢要prepare()必須有問號（？）而不是值。當你做$stmt->execute()這些都考慮到隨後的查詢中。它使您能夠使用不同的值多次執行查詢。

在此之後，執行bind_param()時，您需要提供的數據類型爲每個變量，不只是第一或一般。

下面是一些未經測試的代碼，希望它可以幫助...

$stmt = $mysqli->prepare("INSERT INTO `schauspieler` (
    `name`, `surname`, `about`, `visible`, `admin`, `date`, `file_path`, `photo_type`, `photo_size` 
) VALUES (
    ?, ?, ?, ?, ?, CURRENT_TIMESTAMP, ?, ?, ? 
)"); 

$stmt->bind_param("sssiissi", $name, $surname, $about, $visible, $admin, $file_path, $type, $size); 

有一個嘗試，並告訴我，如果它的工作。

並採取在編制報表的mysqli引用細看;）

http://php.net/manual/de/mysqli.prepare.php