Javascript和PHP - 将值传递给PHP查询 [英] Javascript and PHP - pass value to PHP query

问题描述

亲爱的团队，我是这个网站的新手，也是PHP :)。我需要专家的帮助，我确信我的代码不符合标准，但仍在努力。

我的代码如下所示，我在这里做的是，我从javascript获得了评估ID并将其转换为PHP变量，并且打印了所需的值。

但是当我试图将这个变量放在PHP Query中时，由于我的PHP代码不起作用，它没有传递那些值。但是，如果我传递静态值它工作正常。

Dear Team, I am new on this website and also in PHP :). I need some help from experts, and I am sure my code is not up to the mark but still working on it.
My code is as below, what I do here is, I got evaluation ID from javascript and shift it to a PHP variable and it's printing that required value as well.
but when I am trying to put this variable in PHP Query, it's not passing that values due to which my PHP code is not working. However, if I pass a static value it's working fine.

$sql = 'SELECT * FROM eval where evaid="11075"';

我想把这个$ eid传递给这个提到的查询，但是它没有用。请帮忙。



我尝试过：

I want to pass this $eid into this mentioned query but it's not working. Please help.

What I have tried:

<?php
			$eid = "' + response[0].EvaluationId + '";
			echo $eid;

			$division = "' + response[0].Division + '";
			echo $division;

   	$dbhost = 'localhost'; 
  	 $dbuser = 'root';
  	 $dbpass = '123456';
  	 $conn = mysql_connect($dbhost, $dbuser, $dbpass);
   
 	  if(! $conn ) {
	      die('Could not connect: ' . mysql_error());
 	  }
 	  $sql = 'SELECT * FROM eval where evaid="11075"'; 

 	  mysql_select_db('ops');
 	  $retval = mysql_query( $sql, $conn );
   
 	  if(! $retval ) {
 	     die('Could not get data: ' . mysql_error());
 	  }
   
 		while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {

if ($row['division'] == "CRM Sales - Support" AND $row['etype'] == "dddd")
{
echo "downgrade ";
}
elseif ($row['division'] == "CRM Sales - Support" AND $row['etype'] == "upre")
{ 
echo "update";
}
elseif ($row['division'] == "CRM Sales - Support" AND $row['etype'] == "Ponly")
{
echo "Party";
}
elseif ($row['division'] == "DME")
{
echo "devide me equal";
}
else
{
echo "Open Ended";
}
} 
   mysql_close($conn);

		?>

推荐答案

sql ='SELECT * FROM eval其中evaid =11075';

sql = 'SELECT * FROM eval where evaid="11075"';

我想把这个

I want to pass this

eid传递给这个提到的查询，但它是不工作请帮忙。



我尝试过：

eid into this mentioned query but it's not working. Please help.

What I have tried:

<?php

eid ='+ response [0] .EvaluationId +';
echo

eid = "' + response[0].EvaluationId + '"; echo

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