尝试显示结果时出错(Error When Trying to Display Result)

尝试显示结果时出错(Error When Trying to Display Result)

当我尝试使用PHP显示来自Mysql查询的结果时，我收到以下错误：警告：mysqli_fetch_array（）期望参数1为mysqli_result，第5行的C：\ xampp \ htdocs \ results.php中给出布尔值

这是来自results.php的我的来源：

<?php require 'dbconnect.php'; $q="SELECT Name, Mana Cost, Colour, Set, Ability FROM mtgcards WHERE Name LIKE '%".$_POST['search']."%'"; $r = mysqli_query($dbc, $q); while ($row = mysqli_fetch_array($r)) { echo '<br>'. 'Name: ' . $row['Name'] . ' Mana cost: ' . $row['Mana Cost'] . ' Colour: ' . $row['Colour'] . ' Set: ' . $row['Set'] . ' Ability: ' . $row['Ability']; } mysqli_close($dbc); ?>

我不知道我的问题是什么。 我已经仔细检查了我的表名，并且我已经仔细检查了其他语法问题。 请帮帮我，谢谢！

When I try to display results from a Mysql query with PHP, I get the following error: Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\results.php on line 5

Here's my source from results.php:

<?php require 'dbconnect.php'; $q="SELECT Name, Mana Cost, Colour, Set, Ability FROM mtgcards WHERE Name LIKE '%".$_POST['search']."%'"; $r = mysqli_query($dbc, $q); while ($row = mysqli_fetch_array($r)) { echo '<br>'. 'Name: ' . $row['Name'] . ' Mana cost: ' . $row['Mana Cost'] . ' Colour: ' . $row['Colour'] . ' Set: ' . $row['Set'] . ' Ability: ' . $row['Ability']; } mysqli_close($dbc); ?>

I have no clue what my issue. I have double-checked my table names, and I have double checked other syntax issues. Please help me, thanks!

最满意答案

您需要正确引用包含空格的列名称：

$q="SELECT Name, `Mana Cost`, Colour, Set, Ability FROM mtgcards WHERE Name LIKE '%".$_POST['search']."%'";

You need to properly quote columns names which contain space:

$q="SELECT Name, `Mana Cost`, Colour, Set, Ability FROM mtgcards WHERE Name LIKE '%".$_POST['search']."%'";