dongshuobei1037 2017-02-02 19:21
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查询字符串错误或代码错误？ [重复]

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I am trying to write some code that will simply return a value of someone's name, followed by 3 test scores. In my head, the code should look simply like this:

<html>
<head>
<title> PHP Scores</title>
</head>
<body>
<?php

  echo "<p>",$_GET["name"],"</p>";
  echo "<p>",$_GET["test1"],"</p>";
  echo "<p>",$_GET["test2"],"</p>";
  echo "<p>",$_GET["test3"],"</p>";

?>
</body>
</html>

Now, whenever I type in the URL: http://localhost/phpassignment1.php?name=john&test1=88&test2=74&test3=100

It only returns the value of the name. Is there any reason why I am not getting the 3 test scores to be echoed onto the page?

</div>
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1条回答 默认 最新

  • douxian7117 2017-02-02 19:23
    关注

    Your PHP syntax is wrong for joins. You need to use . instead of ,.

    echo "<p>",$_GET["name"],"</p>";
echo "<p>",$_GET["test1"],"</p>";
echo "<p>",$_GET["test2"],"</p>";
echo "<p>",$_GET["test3"],"</p>";

    Should be:

    echo "<p>" . $_GET["name"] . "</p>";
echo "<p>" . $_GET["test1"] . "</p>";
echo "<p>" . $_GET["test2"] . "</p>";
echo "<p>" . $_GET["test3"] . "</p>";

    Hope this helps!

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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