問題描述

我正在嘗試使用以下代碼(在 $host 中使用適當的值等)對小型 mysql 數據庫進行非常非常簡單的查詢:

I'm trying to make a very, very simple query of a small mysql database, using the following code (with appropriate values in $host, etc.):

$result = mysqli_query($connection, "select university from universities_alpha");
$row = mysqli_fetch_array($result);

echo print_r($result);
echo '<br><br>';
echo print_r($row);

如您所見，我以人類可讀的方式打印了結果，結果是:

As you can see, I printed out the results in a human-readable way, yielding:

mysqli_result Object ( [current_field] => 0 [field_count] => 1 [lengths] => Array ( [0] => 19 ) [num_rows] => 9 [type] => 0 ) 1

Array ( [0] => Arizona State Univ. [university] => Arizona State Univ. ) 1

該專欄中有一些示例大學，所以我不確定我做錯了什么.

There are a few example universities in that column, so I'm not sure what I'm doing wrong.

推薦答案

mysqli_fetch_array 每次調用時都通過指針工作

mysqli_fetch_array works by pointers each time it's called

想象以下內容

$result = mysqli_query($connection, "select university from universities_alpha");
$row = mysqli_fetch_array($result); // this is the first row
$row = mysqli_fetch_array($result); // now it's the second row
$row = mysqli_fetch_array($result); // third row

要以您想要的方式實際顯示數據，我建議您執行以下操作

To actually display the data the way you want it to, I suggest you do the following

$rows = array();
$result = mysqli_query($connection, "select university from universities_alpha");
while($row = mysqli_fetch_array($result)) {
    $rows[] = $row;
}

print_r($rows);

這篇關于mysqli_fetch_array 只返回一個結果的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！