Question

我已经附加了下面的代码。它是从用户界面接受值的表单，我想将结果保存在postgresql数据库中，但我的界面工作正常，但没有插入数据库中的值。

 <?php
 include("connection.php");
 error_reporting(0);
   ?>

 <html>
 <head>
 <title>TESTING INTERFACE</TITLE>

 </head>
 <body>
 <BR>
 TESTING INTERFACE
 <BR>
 <form action ="" method="GET">

   Year <input type="text" name ="Year" value =""/><br><br>
  GDP<input type="text" name ="GDP" value =""/><br><br>
  Gross Indirect Tax <input type="text" name ="Gross_Indirect_Tax" value 
  =""/><br><br>
  CE Revenue<input type="text" name =" CE_Revenue" value =""/><br><br>
  CE Revenue percentage<input type="text" name ="CE_REvenue_Percentage" 
  value =""/><br><br>
  <input type ="submit" name="" value ="submit"/>

  </form>

  <?php
   $yr = $_GET['Year'];
   $gs = $_GET['GDP'];
    $rs1 = $_GET['Gross_Indirect_Tax'];
   $rs2 = $_GET['CE_Revenue'];
   $rs3 = $_GET['CE_REvenue_Percentage'];


 echo $yr;
  echo $gs;
 echo $rs1;
echo $rs2;
echo $rs3;

//echo "$_POST[Year]";
 $query = "INSERT INTO indirecttax 
(Year,GDP,Gross_Indirect_Tax,CE_Revenue,CE_REvenue_Percentage) 
VALUES($yr,$gs,$rs1,$rs2,$rs3)" ;
$x = pg_query($query);
//$qw = pg_query (INSERT INTO indirecttax 
(Year,GDP,Gross_Indirect_Tax,CE_Revenue,CE_REvenue_Percentage) 
 VALUES($yr,$gs,$rs1,$rs2,$rs3)) ;
//$query = "INSERT INTO Criteria_Parameters 
(CriteriaID,DomainID,Criteria_Name,Upper_Limit,Lower_Limit,Range,Unit) 
VALUES 


/*
 $data = pg_query($conn,$qw);

 if($data)
 {
 echo "data inserted into database";

 }
 else
 {
  echo "sorry";
  }
 */
  ?>
  </body>
  </html>

我的connection.php文件工作正常。我的insert语句没有给出结果

Answer 1

您似乎忘了将连接传递给pg_query。

试试这个：

pg_query($conn, $query);

其中$conn是您致电pg_pconnect

所返回的对象

通过php将数据保存到postgres中

1 个答案: