表格内

Question

我试图发布此更早，并认为我会检查回来，等待同行审查，但在我回来它已经走了，我希望我不会不必要地转贴。我在前一篇文章中对我的长期缺席表示歉意。我还在寻找以下行似乎是麻烦的，因为一切最终都会被放入我的数据库：

echo "<td>". 'Tags:<input type="text" name= "MyTag" size="20">'. "</td>"; 

这里有两个文件我用，没有错误消息，只是没有进入MyTag场数据库：

<?php 

    $connect= mysql_connect('localhost', '', ''); 

    if($connect) 
{ 
echo(" 
Connected"); 
} 
else 
{ 
echo(" 
Not Connected"); 
} 

mysql_select_db(''); 

$sql = "select `rebus_tbl2`.`MyOrder`,`rebus_tbl2`.`Word`,`rebus_tbl2`.`Translation`,`rebus_tbl2`.`Picture`,`rebus_t bl2`.`Person`,`rebus_tbl2`.`Picture2`,`rebus_tbl2`.`Object`,`rebus_tbl2`.`Scene`,`rebus_tbl2`.`Tag` from `rebus_tbl2`";  



$query = mysql_query($sql); 

if (!$query) { 
    die(mysql_error()); 
} 



echo "<table border='1'> 

<table bgcolor='#FF7F50'> 


<tr> 
<th>MyOrder</th> 
<th><font color='red'>Word</font></th> 
<th>Translation</th> 
<th>Picture</th> 
<th>Person</th> 
<th>Picture2</th> 
<th>Object</th> 
<th>Scene</th> 
<th>Tag</th> 
<th> Control Panel 
</tr>"; 




while ($row = mysql_fetch_array($query)) { 







    echo "<tr>"; 

    echo "<td>" . $row['MyOrder'] . "</td>"; 
    echo "<td>" . $row['Word'] . "</td>"; 
    echo "<td>" . $row['Translation'] . "</td>"; 
    echo "<td>" ."<img src=\"". $row['Picture'] ."\" alt=\"\" />". "</td>"; 
    echo "<td>" . $row['Person'] . "</td>"; 
    echo "<td>" ."<img src=\"". $row['Picture2'] ."\" alt=\"\" />". "</td>"; 
    echo "<td>" ."<img src=\"". $row['Object'] ."\" alt=\"\" />". "</td>"; 
    echo "<td>" . $row['Scene'] . "</td>"; 
    echo "<td>" . $row['Tag']. "</td>"; 

    echo "<td>". '<form method="post" action="Voting_Submission_Form_to_Voting_Table.php">'."</td>" ; 
    echo "<td>". 'Tags:<input type="text" name= "MyTag" size="20">'. "</td>"; 




     echo "<td>" . '<input type="radio" name = "Likes" Value = "Like" <name> Likes'; 
     echo "<td>" . '<input type="radio" name = "Adds" Value = "Name" <name> Adds'; 
     echo "<td>" . '<input type="radio" name = "Spelling" Value = "Spelling" <name> Spelling'; 
     echo "<td>" . '<input type="radio" name = "Inappropriate" Value = "Inappropriate" <name> Inappropriate'; 


     echo "<td>". '<INPUT TYPE="SUBMIT" VALUE="Submit" NAME="B1">'."</TD>"; 


    echo "</tr>"; 
    } 
    echo "</table>"; 



    ?> 

而且Voting_Submission_Form_to_Voting_Table.php文件

<?php 

$connect= mysql_connect('localhost', '', ''); 


if($connect) 
{ 
echo(" 
Connected"); 
} 
else 
{ 
echo(" 
Not Connected"); 
} 

mysql_select_db(''); 



$MyOrder = $_Post['MyOrder']; 
$MyTag = $_POST['MyTag']; 
$Likes = $_POST['Likes']; 
$Adds = $_POST['Adds']; 
$Spelling = $_POST['Spelling']; 
$Inappropriate = $_POST['Inappropriate']; 


if($_POST) 
{ 
echo(" 
Posted Successfully"); 
} 
else 
{ 
echo(" 
Posted Unsuccessfully"); 
} 

$order = "INSERT INTO Voting_Table 
     (MyOrder, MyTag, Likes, Adds, Spelling, Inappropriate) 
     VALUES 
     ('$MyOrder','$MyTag', '$Likes', '$Adds', '$Spelling', '$Inappropriate')"; 

$result = mysql_query($order); 
if($result) 
{ 
echo(" 
Input data is succeed"); 
} 
else 
{ 
echo(" 
Input data is fail"); 
} 

?> 

Answer 1

你为什么不回应的形式，而不是其他工作表里面在附近。当你在它的时候，你为什么不把所有的东西存储在1个变量中，并在页面上回显一次。

您也没有终止您的最终“<tr>”。它应该是“</tr>”。

此外，虽然它不是你的问题的一部分，请尝试在形成你的html标签时不要使用麻烦的大写和小写字符。坚持使用小写字母。

但是，您的问题可能与您的插入查询或您的数据库连接，你应该已经添加到您的文章，我还没有测试过，可能或可能不会有语法错误，但尝试它，让我知道发生了什么。

<table> 
<form method="post" action="Voting_Submission_Form_to_Voting_Table.php"> 
    <fieldset> 
    <?php 
     $table = "<tr>"; 
     $table .= "<td>" . $row['MyOrder'] . "</td>"; 
     $table .= "<td>" . $row['Word'] . "</td>"; 
     $table .= "<td>" . $row['Translation'] . "</td>"; 
     $table .= "<td>" ."<img src='". $row['Picture'] ."' alt='' />". "</td>"; 
     $table .= "<td>" . $row['Person'] . "</td>"; 
     $table .= "<td>" ."<img src='". $row['Picture2'] ."' alt='' />". "</td>"; 
     $table .= "<td>" ."<img src='". $row['Object'] ."' alt='' />". "</td>"; 
     $table .= "<td>" . $row['Scene'] . "</td>"; 
     $table .= "<td>" . $row['Tag'] . "</td>"; 
     $table .= "<td>" . "<input type='number' name='UserNumber' size='5'/>". "</td>"; 
     $table .= "<td>". "<input type='submit' value='submit' name='b1'>"."</td>"; 
     $table .= "</tr>"; 
     echo $table; 
    ?> 
    </fieldset> 
</form> 
</table>