将 PHP 7 变量插入 SQL Server答案

【问题标题】：Inserting PHP 7 variables into SQL Server将 PHP 7 变量插入 SQL Server
【发布时间】：2019-02-19 18:46:55
【问题描述】：

我确定有人问过这个问题。我已经搜索了答案，但我似乎找不到它。我有这段代码可以将数据从 PHP 连接到 MS Sql 数据库中。它不工作。

 <!DOCTYPE html>
<!--
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and open the template in the editor.
-->
<?php
//Session Start
session_start();
include '/includes/dbconn.mssql.php';

ini_set('display_errors','On');
?>
<html>
<head>
    <meta charset="UTF-8">
    <title></title>
</head>
<body>
    <div>

    </div>
</body>
<?php        

        $sql = "INSERT INTO 
agents(fname,lname,date,dl1,ul1,lat1,jit1,dl2,ul2,lat2,jit2,dl3,ul3,lat3,jit3,UCN,ISP,IP,mmake,mmodel,[user],pass,email,winver,modem,mon1mk,mon1mdl,mon2mk,mon2mdl,headset,hmake,hmodel,webcam,prospect,[current],planul,plandl,station) "
                    . "VALUES('Janice','Smith','2/15/2019','10','10','1','1','10','10','1','1','10','10','1','1','333689','Xfinity','10.0.0.1','Ubee','100','admin','password','jane.smith@companyinc.com','null','null','Acer','100','AOC','5','1','Plantronic','54','1','0','1','150','50','28')";
            //      $sql = "INSERT INTO agents(fname,lname,date,dl1,ul1,lat1,jit1,dl2,ul2,lat2,jit2,dl3,ul3,lat3,jit3,UCN,ISP,IP,mmake,mmodel,user,pass,email,winver,modem,mon1mk,mon1mdl,mon2mk,mon2mdl,workspace,monitor1,monitor2,headset,hmake,hmodel,webcam,prospect,current,planul,plandl,station) VALUES('$fname','$lname','$date','$dl1','$ul1','$lat1','$jit1','$dl2','$ul2','$lat2','$jit2','$dl3','$ul3','$lat3','$jit3','$UCN','$ISP','$IP','$mmake','$mmodel','$user','$pass','$email','$winver','$modem','$mon1mk','$mon1mdl','$mon2mk','$mon2mdl','$workspace','$monitor1','$monitor2','$headset','$hmake','$hmodel','$webcam','$prospect','$current','$planul','$plandl','$station')";
            $result = sqlsrv_query($conn,$sql);
            $echo = sqlsrv_rows_affected;
            if(!$result) {
    echo 'Your code failed. $echo';
}
else {
    echo 'Success! $echo'; 
}
?>
</html>

当我在 Edge 中运行代码时，它会显示

致命错误：在 .... 第 34 行调用未定义函数 sqlsrv_query()

第 34 行是

$result = sqlsrv_query($conn,$sql);

我做错了什么？

【问题讨论】：

您没有任何与undefined function sqlsrv_connect 相关的错误吗？你应该在 sqlsrv_query 之前得到它。确保 sqlsrv 扩展 is installed
这也很可疑：include '/includes/dbconn.mssql.php'; 因为它暗示includes 目录位于文件系统根目录，而不是相对于 PHP 中的其他目录。您的系统中可能就是这种情况，但似乎不太可能并且非常罕见。并且可以解释为什么 no 没有从 sqlsrv_connect() 得到先前的错误。
你已经打开了 display_errors 这很好，也就是在该行之前设置 error_reporting(E_ALL);

标签： php html sql-server database

【解决方案1】：

我可以将数据插入到带有数据和变量的单元格中。

     <?php

   $uploadDir = 'pics/';

   if(isset($_POST['upload'])){ 

       $fileName = $_FILES['monitor1']['name'];
       $tmpName  = $_FILES['monitor1']['tmp_name'];
       $fileSize = $_FILES['monitor1']['size'];
       $fileType = $_FILES['monitor1']['type'];

       $filePath = $uploadDir . $fileName;

       $result = move_uploaded_file($tmpName, $filePath);

       if (!$result) {
           echo "Error uploading <strong>file</strong>";
           exit;
}


       $sql =  "INSERT INTO agents(fname,lname,employee_status,dl1,ul1,lat1,jit1,monitor1) "
               . "  VALUES('".$_POST['fname']."',
                        '" .$_POST['lname']."',
                        '" .$_POST['status']."',
                        '" .$_POST['dl1']."',
                        '" .$_POST['ul1']."',
                        '" .$_POST['lat1']."',
                        '" .$_POST['jit1']."',
                        '" .$_FILES['monitor1']."')";                            

              $result = sqlsrv_query($conn,$sql);

              if( $result === false ) {
 die( print_r( sqlsrv_errors(), true));
}

   }

?>

【讨论】：