duanpu5048 2015-04-26 10:05
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如何使用HTML链接将变量从一个PHP文件传递到另一个PHP文件?

    //DB CONNECTION
   $sql = "SELECT `city`,`country` from infotab";
   $result = $conn->query($sql);

   while ($row = $result->fetch_assoc()) {
     echo  $row["city"].$row["country"]"<a href='order.php'>order</a>"; }

Table output:

http://imgur.com/jH2VzBf

This code will select data. Additionally, there is reference to order.php on every line. When the user clicks on reference( <a href> clause), it opens order.php and there I need to know which row the user selected to work with these data.

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  • dougongnan2167 2015-04-26 10:14
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    Change the code to:

    while ($row = $result->fetch_assoc()) {         
        echo  $row["city"] . $row["country"] . "<a href='order.php?city=" . $row["city"] . "&country=" . $row["country"] . "'>order</a>";
    }
    

    In order.php you can then access these values by using the $_GET["city"] and $_GET["country"] variables which contain the values from your <a href> link on the previous page. For example, running echo $_GET["city"]; will output the city name.

    Edit: As @Rizier123 pointed out, using a unique ID might be more prone to errors in case your database contains more than one entry for the same city or country. You should consider introducing an ID in your table structure and then using that in the link to order.php.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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