我有这样的代码。
代码:
<?php
$book_query = mysql_query("select * from book_master')");
while($book_query_fetch = mysql_fetch_assoc($book_query)){
echo "<pre>";
print_r($book_query_fetch);
echo "</pre>"
}
?>
输出:
Array
(
[Book_Name] => Book1
[Book_ID] => 123
)
Array
(
[Book_Name] => Book2
[Book_ID] => 124
)
预期输出:(在表格中)
Book Name Book_ID
Book1 123
Book2 124
我怎样才能实现这个目标? 编辑: header 部分是动态负载。所以我也需要循环中的表头
最佳答案
我不知道你在哪里坚持这样做,但你可以在下面这样做,
echo "<table>";
$i = 0;
while($row = mysql_fetch_assoc($book_query))
{
if($i == 0){
$columns = array_keys($row);
echo "<th>";
foreach($columns as $column){
echo "<td> $column</td>";
}
echo "</th>";
}
echo'<tr>';
echo '<td>'.$row['Book_Name'].'</td>';
echo '<td>'.$row['Book_ID'].'</td>';
echo '</tr>';
$i++;
}
echo "</table>";
华林: Please, don't use mysql_*
functions in new code 。它们不再维护and are officially deprecated 。请参阅red box ?了解 prepared statements相反,并使用 PDO ,或MySQLi -this article将帮助您决定哪个。如果您选择 PDO,here is a good tutorial .
关于php - Mysql fetch与PHP MYSQL以表格式关联,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22784349/