Mysql查詢沒有將數據插入到數據庫中

我試圖創建一個連接到數據庫的網站。我的Select語句用於顯示數據，但我的插入和我的刪除語句實際上都不會改變數據。我搜索了谷歌和stackoverflow，並試圖嘗試各種「答案」，當我陷入困境，但它仍然無法正常工作。結果，我的代碼的currect狀態是可能冗餘代碼的大雜燴。一般來說，我對HTML和PHP相當陌生，所以我會很感激這個幫助。我也道歉，如果我的格式是錯誤的，因爲我從來沒有使用過像stackoverflow之前。以下是該網頁的一個版本，供那些不會在實際中看到問題的人使用。編輯：修正。每個人的評論和回答都很有幫助，但事實證明，我有一個if語句來檢查以確保輸入只是字母，當前兩個變量包含數字時。Mysql查詢沒有將數據插入到數據庫中

如果（的preg_match（「/ [AZ | AZ] + /」，$ _ POST [ '名']））是罪魁禍首

<!DOCTYPE html> 
<html> 
<head> 
<link rel = "stylesheet" type = "text/css" href = "style.css" /> 
</head> 
<body> 
    <div> 
    <form method="post" action="DBADELETE.php?go" id="deleteform"> 
     <input name="GMOVIE" type="text" placeholder="GMOVIEID of a movie to delete" required=""> 
    <input name="UMOVIE" type="text" placeholder="UMOVIEID of a movie to delete" required=""> 
     <input type="submit" name="submit" value="DELETE">    
     </form> 
    </div> 
<div> 
    <form method="post" action="DBAUPDATE.php?go" id="updateform"> 
     <input name="GMOVIE" type="text" placeholder="GMOVIEID of a movie to update" required=""> 
    <input name="UMOVIE" type="text" placeholder="UMOVIEID of a movie to update" required=""> 
<input name="TITLE" type="text" placeholder="TITLE of a movie to update" required=""> 
<input name="GENRE" type="text" placeholder="GENRE of a movie to update" required=""> 
<input name="RATING" type="text" placeholder="RATING of a movie to update" required=""> 
     <input type="submit" name="submit" value="UPDATE">    
     </form> 
    </div> 
    <div id = "content1" align = "center"> 
    <table id = "MOVIES"> 
    <tr> 
     <th>GMOVIEID</th> 
     <th>UMOVIEID</th> 
     <th>TITLE</th> 
     <th>GENRE</th> 
     <th>RATING</th> 
    </tr> 
<?php 
include 'bd.php'; 
$conn = new mysqli($host, $user, $pw, $bd_name); 

if ($conn->connect_error) { 
    die("Connection Failed: " . $conn->connect_error); 
} 
if(isset($_POST['submit'])){ 
$gmovie=$_POST['GMOVIE']; 
$umovie=$_POST['UMOVIE']; 
$title=$_POST['TITLE']; 
$genre=$_POST['GENRE']; 
$rating=$_POST['RATING']; 
$gmov_safe = mysqli_real_escape_string($gmovie,$conn); 
$umov_safe = mysqli_real_escape_string($umov,$conn); 
$title_safe = mysqli_real_escape_string($title,$conn); 
$genre_safe = mysqli_real_escape_string($genre,$conn); 
$rating_safe = mysqli_real_escape_string($rating,$conn); 
$q ="INSERT INTO MOVIE (GMOVIEID, UMOVIEID, TITLE,GENRE, RATING)  VALUES($gmovie,$umovie,'$title_safe','$genre_safe','$rating_safe')" 
$conn->query($q); 

if ($conn->query($q) === TRUE) 
{ 
echo "New record created successfully"; 
} 
else 
{ 
echo "Error: " . $q . "<br>" . $conn->error; 
} 
} 
?> 
<?php 
include 'bd.php'; 
$conn = new mysqli($host, $user, $pw, $bd_name); 

if ($conn->connect_error) { 
    die("Connection Failed: " . $conn->connect_error); 
} 
$q = "SELECT GMOVIEID, UMOVIEID, TITLE, GENRE, RATING FROM MOVIE"; 
$result = $conn->query($q); 

$row_count = $result->num_rows; 

while ($row = $result->fetch_assoc()) { 
    if (($row_count % 2) == 1) { 
     echo "<tr>"; 
    } else { 
     echo "<tr class=\"alt\">"; 
    } 

    echo "<td>" . $row[GMOVIEID] . "</td>" . 
     "<td>" . $row[UMOVIEID] . "</td>" . 
     "<td>" . $row[TITLE] . "</td>" . 
     "<td>" . $row[GENRE] . "</td>" . 
     "<td>" . $row[RATING] . "</td>" . 
     "</tr>" . PHP_EOL; 

    $row_count--; 
    } 
?> 
</body> 
<?php $conn->close(); ?> 
</html>

來源

2016-12-05 F Perrault

經典的MySQL API和語法錯誤 –

混合不健全的愚蠢，但到底是什麼意思呢？ –

我首先看到的可能是這個問題：if（isset（$ _ GET ['go']））{。這是一個POST請求，所以$ _GET不會被設置，在這種情況下不會觸發其他任何事情。 – princessjackie

問題似乎是形式重新路由到另一個在頁面提交：

action="DBAUPDATE.php?go"

或

action="DBADELETE.php?go"

但樣板代碼一起插入語句是在當前頁面：DBAMOVIES.php

你需要添加插入語句和DBAUPDATE.php和DBADELETE.php樣板代碼。

只需創建上述兩個頁面，並添加以下語句：

<html> 
<body> 
    <?php include("DBAMOVIES.php"); ?> 
</body> 
</html>

來源

2016-12-06 02:55:41

Mysql查詢沒有將數據插入到數據庫中

回答

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