Question

我花了几天时间试图计算如何在我的网页上获取表单数据以插入到forklift数据库mssql上的employee表中。当我单击表单上的提交时，它会刷新页面并建立连接，但数据库中没有数据。

<?php

/* Connect using Windows Authentication. */

$serverName = "EXAMPLE";
$connectionOptions = array("Database"=>"FORKLIFT");

/* Connect using Windows Authentication. */
$conn = sqlsrv_connect( $serverName, $connectionOptions);

if($conn) 

{
echo "Connection established.<br />";
}

else 

{
echo "Connection could not be established.<br />";
die(print_r(sqlsrv_errors(), true));

}

if(empty($_POST) === false && empty($errors)=== true)

{ 
//assign form input to variables
$FIRSTNAME = $_POST["FIRSTNAME"];
$LASTNAME = $_POST["LASTNAME"];
$DATEOFBIRTH = $_POST["DATEOFBIRTH"];
$PHONENUMBER = $_POST["PHONENUMBER"];
$ADDRESS = $_POST["ADDRESS"];

/*Insert data.*/
$INSERT_ROW = $query = "INSERT INTO              
EMPLOYEE(FIRSTNAME,LASTNAME,DATEOFBIRTH,PHONENUMBER,ADDRESS)
VALUES ('$FIRSTNAME','$LASTNAME','$DATEOFBIRTH','$PHONENUMBER','$ADDRESS')";

$result = sqlsrv_prepare($conn,$query)or die('Error querying MSSQL     
database');
        sqlsrv_execute($result);              
}

?>

HTML

        <form name="submit" action="employee.php" method="POST" >


           <h2>Register New Member</h2>

           <table border="0">
              <tr>
                 <td>FIRSTNAME</td>
                 <td>
                    <input type="text" name="FIRSTNAME" id="FIRSTNAME"/>
                 </td>
              </tr>

              <tr>
                 <td>LASTNAME</td>
                 <td>
                    <input type="text" name="LASTNAME" id="LASTNAME"/>
                 </td>
              </tr>

              <tr>
                 <td>DATE_OF_BIRTH</td>
                 <td>
                    <input type="date" name="DATE_OF_BIRTH" id="DATE_OF_BIRTH"/>
                 </td>
              </tr>   

              <tr>
                 <td>PHONENUMBER</td>
                 <td>
                    <input type="text" name="PHONENUMBER" id="PHONENUMBER"/>
                 </td>
              </tr>     

              <tr>
                 <td>ADDRESS</td>
                 <td>
                    <input type="text" name="ADDRESS" id="ADDRESS"/>
                 </td>
              </tr>  


              <tr>
                 <td></td>
                 <td align="right"><input type="submit" name="submit" value="REGISTER"/></td>

              </tr>
           </table>

Answer 1

我进行了一些小的改动，但无法测试看 - 运行它，看看结果sql语句是什么样的。我希望它有所帮助。

<?php
    if( !empty( $_POST ) && isset( $_POST["FIRSTNAME"], $_POST["LASTNAME"], $_POST["DATE_OF_BIRTH"], $_POST["PHONENUMBER"], $_POST["ADDRESS"] ) ) {

        /* 
            There is no point creating a db connection if a later condition fails
            so create the db conn after testing key variables are set
        */

        $server = "EXAMPLE";
        $options = array( "Database"=>"FORKLIFT" );


        $conn = sqlsrv_connect( $server, $options );

        /* In production environment do not display actual errors! */
        if( !$conn ) die( print_r( sqlsrv_errors(), true ) );   


        /* as pointed out by @Daniel Rutter, the field is actually called `DATE_OF_BIRTH` */
        $FIRSTNAME = $_POST["FIRSTNAME"];
        $LASTNAME = $_POST["LASTNAME"];
        $DATEOFBIRTH = $_POST["DATE_OF_BIRTH"];
        $PHONENUMBER = $_POST["PHONENUMBER"];
        $ADDRESS = $_POST["ADDRESS"];



        $sql = "INSERT INTO EMPLOYEE
            ( FIRSTNAME,LASTNAME,DATEOFBIRTH,PHONENUMBER,ADDRESS )
                VALUES
            ( '$FIRSTNAME','$LASTNAME','$DATEOFBIRTH','$PHONENUMBER','$ADDRESS' )";



        /* Comment out the following line after running the script and observing the output here */
        exit( $sql );


        /* I assume that `sqlsrv_prepare` returns either true or false */
        $stmt = sqlsrv_prepare( $conn, $sql ) or die( 'Error querying MSSQL database' );

        if( $stmt ) sqlsrv_execute( $stmt );

    /* debug post vars */           
    } else { echo 'Error::' . print_r($_POST,true ); }
?>

简要介绍一下各种sqlsrv_*命令的在线PHP手册，或许你应该尝试这些方法 - 它与给出的一个例子紧密相关 - 尽管它没有经过测试，因为没有要使用的mssql服务器实例。

<?php
    if( !empty( $_POST ) && isset( $_POST["FIRSTNAME"], $_POST["LASTNAME"], $_POST["DATE_OF_BIRTH"], $_POST["PHONENUMBER"], $_POST["ADDRESS"] ) ) {

        /* 
            There is no point creating a db connection if a later condition fails
            so create the db conn after testing key variables are set
        */

        $server = "EXAMPLE";
        $options = array( "Database"=>"FORKLIFT" );


        $conn = sqlsrv_connect( $server, $options );

        /* In production environment do not display actual errors! */
        if( !$conn ) die( print_r( sqlsrv_errors(), true ) );   



        $FIRSTNAME = $_POST["FIRSTNAME"];
        $LASTNAME = $_POST["LASTNAME"];
        $DATEOFBIRTH = $_POST["DATE_OF_BIRTH"];
        $PHONENUMBER = $_POST["PHONENUMBER"];
        $ADDRESS = $_POST["ADDRESS"];



        $sql = "INSERT INTO EMPLOYEE
            ( FIRSTNAME, LASTNAME, DATEOFBIRTH, PHONENUMBER, ADDRESS )
                VALUES
            ( ?,?,?,?,? )";

        $params=array(
            &$FIRSTNAME,
            &$LASTNAME,
            &$DATEOFBIRTH,
            &$PHONENUMBER,
            &$ADDRESS
        );

        /* add the posted variables as an array here as the third arg */
        $stmt = sqlsrv_prepare( $conn, $sql, $params ) or die( 'Error querying MSSQL database' );
        if( $stmt ) sqlsrv_execute( $stmt );



    /* debug post vars */           
    } else { 
        echo 'Error::' . print_r( $_POST, true );
    }
?>

关于sqlsrc_prepare的PHP在线手册：

准备执行查询。此功能非常适合准备将使用不同参数多次执行的查询值。

和

当您准备一个使用变量作为参数的语句时，变量绑定到语句。这意味着如果您更新下次执行语句时变量的值它将使用更新的参数值运行。对于你的陈述计划只执行一次，使用sqlsrv_query（）。

Answer 2

我注意到你试图将数据发布到数据库，这实际上并没有在PHP端正确命名，这只会导致与你的DATE OF BIRTH相关的信息首先没有发布，因为它当您声明要检索发布的数据时，必须完全匹配PHP方面。我非常肯定，从内存中你只需要将客户端表单中的名称更改为DATEOFBIRTH，否则你必须将PHP端更改为DATE_OF_BIRTH而不是此，同时将客户端保持为DATE_OF_BIRTH

<input type="date" name="DATE_OF_BIRTH" id="DATE_OF_BIRTH"/>

需要：

<input type="date" name="DATEOFBIRTH" id="DATEOFBIRTH"/>

网站表单没有向mssql数据库提交数据

2 个答案: