Question

如何使这种情况在php中起作用？

不工作的情况（我要工作的情况）：我有一个表，用于存储要在我的php代码中使用的where子句。

它作为字符串存储在表的字段"WHERECLAUSE"中：

"where WEEK = '$weeknum'  AND YEAR = '$year'"

我从表字段“ WHERECLAUSE”中选择一个字符串值并放入变量

php代码：

   $TABLE = 'REASSORT';
   $whereclause =  "where WEEK = '$weeknum'  AND YEAR = '$year'";

     // create the select query
   $query = "select * FROM $TABLE " . $whereclause;

当我echo $query收到->

 select * FROM REASSORT where WEEK = '$weeknum'  AND YEAR = '$year'

由于变量$weeknum和$year呈现为字符串，因此它不返回表“ REASSORT”中的数据。

我希望的工作情况：

PHP代码：

$whereclause = " where WEEK = '$weeknum'  AND YEAR = '$year' ";
$query = "select * FROM $TABLE " . $whereclause;

当我回显$ query时，我得到->

select * FROM REASSORT where WEEK = '4' AND YEAR = '2019'

并且数据从表“ REASSORT”返回

预先感谢您的帮助。

Answer 1

尝试将$weeknum和$year放在双引号（“）而不是单引号（'）中，如下所示

$whereclause = 'where WEEK = "$weeknum" AND YEAR = "$year";

作为单引号下的变量，将被视为字符串而不是变量。

Answer 2

请用双引号（“）替换单引号（'）

替换

$whereclause = " where WEEK = '$weeknum' AND YEAR = '$year' ";

使用

$whereclause = ' where WEEK = "$weeknum" AND YEAR = "$year" ';

Answer 3

PHP无法将'$weeknum'和'$year'识别为变量。您应该尝试使用eval()，但请谨慎使用此功能。

$weeknum = 5;
$year = 2019;
$whereclause = "where WEEK = '$weeknum' AND YEAR = '$year'";

eval("\$whereclause = \"$whereclause\";");

$query = "select * FROM $TABLE " . $whereclause;

$ query然后应输出：

"select * FROM REASSORT where WEEK = '5'  AND YEAR = '2019'"

或使用花括号代替单引号。

所以代替： '$weeknum'和'$year'

应该是 {$weeknum}和{$year}

有关eval()的详细信息，请参阅： http://php.net/manual/en/function.eval.php

有关使用大括号的信息，请参见： https://www.phpknowhow.com/basics/usage-of-brackets/

Answer 4

您需要使用eval来强制PHP将变量的值替换为$whereclause字符串。但请注意，如果您不知道值可能是eval，可能会很危险。这将起作用：

$query = "select * FROM $TABLE " . eval("return \"$whereclause\";");
echo $query;

输出：

select * FROM REASSORT where WEEK = '4' AND YEAR = '2019'

Demo on 3v4l.org

如何在字符串中输出PHP变量值以用作MySQL查询的一部分？

4 个答案: