php mysqli_fetch_array while 循环列
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14456529/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
mysqli_fetch_array while loop columns
提问by Sebastian
Should be pretty basic, but I can't get it to work. I have this code to iterate over a mysqli query:
应该是非常基本的，但我无法让它工作。我有这个代码来迭代一个 mysqli 查询：
while($row = mysqli_fetch_array($result)) {
$posts[] = $row['post_id'].$row['post_title'].$row['content'];
}
It works and returns:
它工作并返回：
Variable #1: (Array, 3 elements) ? 0 (String): "4testtest" (9 characters) 1 (String): "1Hello world!Welcome to WordPress. This is your first post. Edit or delete it, then start blogging!" (99 characters) 2 (String): "2Sample PageThis is an example page. It's different from a blog post because it will stay in one place and will show up in your site navigation (in most themes)." (161 characters)
变量#1：（数组，3 个元素）？0（字符串）：“4testtest”（9 个字符） 1（字符串）：“1Hello world！欢迎来到 WordPress。这是您的第一篇文章。编辑或删除它，然后开始写博客！” （99 个字符）2（字符串）：“2Sample PageThis 是一个示例页面。它与博客文章不同，因为它会停留在一个地方，并会显示在您的网站导航中（在大多数主题中）。” （161 个字符）
The problem is that it puts all three colums into one column, so I can't access them seperatly.
问题是它将所有三个列都放在一列中，所以我无法单独访问它们。
This for example:
这例如：
0 (String): "4testtest" (9 characters)
0（字符串）：“4testtest”（9 个字符）
Should be seperated into 4, test, test
应该分为4个，测试，测试
When I do this:
当我这样做时：
while($row = mysqli_fetch_array($result)) {
$posts['post_id'] = $row['post_id'];
$posts['post_title'] = $row['post_title'];
$posts['type'] = $row['type'];
$posts['author'] = $row['author'];
}
It only outputs 1 row instead of all three …
它只输出 1 行而不是所有三行……
Any help is greatly appreciated!
任何帮助是极大的赞赏！
回答by Ricardo Ortega Maga?a
Get all the values from MySQL:
从 MySQL 获取所有值：
$post = array();
while($row = mysql_fetch_assoc($result))
{
$posts[] = $row;
}
Then, to get each value:
然后，获取每个值：
<?php
foreach ($posts as $row)
{
foreach ($row as $element)
{
echo $element."<br>";
}
}
?>
To echo the values. Or get each element from the $post variable
呼应这些值。或者从 $post 变量中获取每个元素
回答by Bogdan Lozyak
This one was your solution.
这是您的解决方案。
$x = 0;
while($row = mysqli_fetch_array($result)) {
$posts[$x]['post_id'] = $row['post_id'];
$posts[$x]['post_title'] = $row['post_title'];
$posts[$x]['type'] = $row['type'];
$posts[$x]['author'] = $row['author'];
$x++;
}
回答by 05beckga
I think this would be a more simpler way of outputting your results.
我认为这将是一种更简单的输出结果的方式。
Sorry for using my own data should be easy to replace .
抱歉使用我自己的数据应该很容易替换。
$query = "SELECT * FROM category ";
$result = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($result))
{
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
echo $cat_id . " " . $cat_title ."<br>";
}
This would output :
这将输出：
- -ID Title
- -1 Gary
- -2 John
- -3 Michaels
- -身标题
- -1 加里
- -2 约翰
- -3 迈克尔斯
回答by Elangovan
Both will works perfectly in mysqli_fetch_arrayin while loops
两者都可以在 while 循环中的mysqli_fetch_array中完美运行
while($row = mysqli_fetch_array($result,MYSQLI_BOTH)) {
$posts[] = $row['post_id'].$row['post_title'].$row['content'];
}
(OR)
（或者）
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) {
$posts[] = $row['post_id'].$row['post_title'].$row['content'];
}
mysqli_fetch_array()- has second argument $resulttype.
mysqli_fetch_array()- 有第二个参数 $resulttype。
MYSQLI_ASSOC:Fetch associative array
MYSQLI_ASSOC：获取关联数组
MYSQLI_NUM:Fetch numeric array
MYSQLI_NUM：获取数值数组
MYSQLI_BOTH:Fetch both associative and numeric array.
MYSQLI_BOTH：获取关联数组和数值数组。
回答by Fero
Try this :
尝试这个 ：
$i = 0;
while($row = mysqli_fetch_array($result)) {
$posts['post_id'] = $row[$i]['post_id'];
$posts['post_title'] = $row[$i]['post_title'];
$posts['type'] = $row[$i]['type'];
$posts['author'] = $row[$i]['author'];
}
$i++;
}
print_r($posts);
回答by Pepito Rosales
Try this...
尝试这个...
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
