插入-PHP& SQL服务器 [英] INSERT - PHP & SQL Server

问题描述

我不知道发生了什么，但是它只是不想工作.

I don't know what is going on, but it just doesn't want to work.

提交表单时，我一直收到此错误:

I keep getting this error when I submit my form:

Array([0] => Array([0] => 22001 [SQLSTATE] => 22001 [1] => 8152 [code] => 8152 [2] => [Microsoft] [SQL Server本机客户端10.0 ] [SQL Server]字符串或二进制数据将被截断.[消息] => [Microsoft] [SQL Server Native Client 10.0] [SQL Server]字符串或二进制数据将被截断.)[1] =>数组([0 ] => 01000 [SQLSTATE] => 01000 [1] => 3621 [code] => 3621 [2] => [Microsoft] [SQL Server Native Client 10.0] [SQL Server]该语句已终止. => [Microsoft] [SQL Server Native Client 10.0] [SQL Server]该语句已终止.

Array ( [0] => Array ( [0] => 22001 [SQLSTATE] => 22001 [1] => 8152 [code] => 8152 [2] => [Microsoft][SQL Server Native Client 10.0][SQL Server]String or binary data would be truncated. [message] => [Microsoft][SQL Server Native Client 10.0][SQL Server]String or binary data would be truncated. ) [1] => Array ( [0] => 01000 [SQLSTATE] => 01000 [1] => 3621 [code] => 3621 [2] => [Microsoft][SQL Server Native Client 10.0][SQL Server]The statement has been terminated. [message] => [Microsoft][SQL Server Native Client 10.0][SQL Server]The statement has been terminated. ) )

这是 PHP 代码:

    <?php
$who = $_REQUEST["who"];
$what = $_REQUEST["what"];

$serverName = "xxx";   
$uid = "xxx";     
$pwd = "xxx";    
$databaseName = "xxx";   

$connectionInfo = array( "UID"=>$uid,                              
                         "PWD"=>$pwd,                              
                         "Database"=>$databaseName);   

/* Connect using SQL Server Authentication. */    
$conn = sqlsrv_connect( $serverName, $connectionInfo);    

$tsql = "insert into Suggestions (Who, What, Votes) values ('$who','$what','10')";   

/* Execute the query. */    

$stmt = sqlsrv_query( $conn, $tsql);    

if ( $stmt )    
{    
     $something = "Submission successful.";
}     
else     
{    
     $something = "Submission unsuccessful.";
     die( print_r( sqlsrv_errors(), true));    
}
    $output=$something;
/* Free statement and connection resources. */    
sqlsrv_free_stmt( $stmt);    
sqlsrv_close( $conn);
?>

这是HTML表单:

<form action="startvoting.php" method="post" id="myform">
          <ol>
            <li>
              <label for="name">Nickname</label>
              <input id="who" name="who" class="text" />
            </li>
            <li>
              <label for="message">What <strong>you</strong> Want</label>
              <textarea id="what" name="what"></textarea>
            </li>
            <li class="buttons">
              <input type="image" src="images/send.gif" class="send" />
              <div class="clr"></div>
            </li>
          </ol>
        </form>

有人可以帮我吗?我不知道该怎么办！

Can someone please help me? I don't know what to do!

谢谢

更新

这是定义:

TABLE_QUALIFIER TABLE_OWNER TABLE_NAME  COLUMN_NAME DATA_TYPE   TYPE_NAME   PRECISION   LENGTH  SCALE   RADIX   NULLABLE    REMARKS COLUMN_DEF  SQL_DATA_TYPE   SQL_DATETIME_SUB    CHAR_OCTET_LENGTH   ORDINAL_POSITION    IS_NULLABLE SS_DATA_TYPE
DB_11967_suggestions    dbo Suggestions Who 12  varchar 1   1           1           12      1   1   YES 39
DB_11967_suggestions    dbo Suggestions What    12  varchar 1   1           1           12      1   2   YES 39
DB_11967_suggestions    dbo Suggestions Votes   4   int 10  4   0   10  1           4           3   YES 38

抱歉，它的格式不正确.

Sorry it's not properly formatted.

推荐答案

我认为您在column(fields)类型中存在错误，请尝试仅插入一个字符，然后提交成功，然后尝试扩展..即增加字段char num ... etc

I think you have an error in columns(fields) types , try insert just one character , then the submission successfully , try expand fields type .. , i.e. increase char num ...etc

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