PHP圖像上傳並將您的路徑插入數據庫

Question

我有一個文件輸入的錯誤。確實，圖像被移動到服務器的文件夾 - isset（$ _ FILES ['avatar']），沒問題 - 但user_avatar的字段沒有填寫你的名字。它的數據庫名稱爲「default_130x130.png」。我應該如何將「$ user_avatar」變量從upload_image.php發送到new_user.php？

---

EN /用戶/ new.html

<form id="avatar_file_upload_form" action="../../utils/upload_image.php" method="post" enctype='multipart/form-data'style="position:absolute;z-index:-10;height:1px;width:1px;overflow:hidden;visibility:hidden;"> 
     <input type="file" name="avatar" id="avatar_file_upload_field" accept="image/jpeg,image/pjpeg,image/bmp,image/gif,image/jpeg,image/png"/> 
     <input type="submit" /> 
    </form> 

---

utils的/ upload_image.php

<?php 
ob_start(); 
$mimeExt = array(); 
$mimeExt['image/jpeg'] ='.jpg'; 
$mimeExt['image/pjpeg'] ='.jpg'; 
$mimeExt['image/bmp'] ='.bmp'; 
$mimeExt['image/gif'] ='.gif'; 
$mimeExt['image/x-icon'] ='.ico'; 
$mimeExt['image/png'] ='.png'; 
if(isset($_FILES["avatar"])) { 
    //Begins image upload 
     $user_avatar = md5(uniqid(time())).$mimeExt[$_FILES["avatar"]["type"]]; //Get image extension 
     $user_avatar_dir = "../img/".$user_avatar; //Path file 
     move_uploaded_file($_FILES["avatar"]["tmp_name"], $user_avatar_dir); 

} else { 
    $user_avatar = "default_130x130.png"; 
} 
?> 

---

new_user.php

<?php 
ob_start(); 
include "config.php"; 
include "utils/upload_image.php"; 
$sql = mysql_query("insert into user(user_avatar) values('$user_avatar')", $db_connection) or die("Error: ".mysql_Error()); 
ob_end_clean(); 
mysql_close($db_connection); 
?> 

---

注意：「config.php」文件工作正常。

Answer 1

我認爲你應該將表單的動作屬性設置爲new_user.php。現在你將它路由到upload_image.php，我不知道你在那之後如何加載new_user.php。當沒有文件發送到該頁面時，實際上包含upload_image.php的new_user.php。

這應該是你的HTML：

<form id="avatar_file_upload_form" action="new_user.php" method="post" enctype='multipart/form-data'style="position:absolute;z-index:-10;height:1px;width:1px;overflow:hidden;visibility:hidden;"> 
    <input type="file" name="avatar" id="avatar_file_upload_field" accept="image/jpeg,image/pjpeg,image/bmp,image/gif,image/jpeg,image/png"/> 
    <input type="submit" /> 
</form> 

這應該是你upload_image.php：

<?php 
$mimeExt = array(); 
$mimeExt['image/jpeg'] ='.jpg'; 
$mimeExt['image/pjpeg'] ='.jpg'; 
$mimeExt['image/bmp'] ='.bmp'; 
$mimeExt['image/gif'] ='.gif'; 
$mimeExt['image/x-icon'] ='.ico'; 
$mimeExt['image/png'] ='.png'; 
if(isset($_FILES["avatar"])) { 
    //Begins image upload 
     $user_avatar = md5(uniqid(time())).$mimeExt[$_FILES["avatar"]["type"]]; //Get image extension 
     $user_avatar_dir = "../img/".$user_avatar; //Path file 
     move_uploaded_file($_FILES["avatar"]["tmp_name"], $user_avatar_dir); 

} else { 
    $user_avatar = "default_130x130.png"; 
} 
?> 

，這應該是你的new_user.php：

<?php 
ob_start(); 
include "config.php"; 
include "utils/upload_image.php"; 
$sql = mysql_query("insert into user(user_avatar) values('$user_avatar')", $db_connection) or die("Error: ".mysql_Error()); 
ob_end_clean(); 
mysql_close($db_connection); 
?> 

Answer 2

形式的行動是utils/upload_image.php的代碼和流程說你需要設置表單acti登錄到new_user.php文件。

而且我不知道爲什麼，因爲到utils/upload_image.php頁面進行提交插入查詢工作，並沒有任何關係可見之間utils/upload_image.php到new_user.php ..你插入查詢不應該在這種情況下

更新工作：爲

我該如何做到真正的插入，所以？

讓你的形式去new_user.php改變action="../../new_user.php"

<form id="avatar_file_upload_form" action="../../new_user.php" method="post" enctype='multipart/form-data'style="..."> 

</form>