[英]Only one row returned in mysql data retrieval
首先,对这个漫长的问题感到抱歉...但是,这正在引起我的注意。任何帮助都会被我们接受。
我编写了以下函数以从mysql数据库返回数据:
function injuryTable()
{
# get all players that are injured and their injuires...
$sql = "SELECT players.id, players.injury_id, players.pname, injuries_date.name, start_date, end_date
FROM players INNER JOIN injuries_date
ON injury_id = injuries_date.id";
$result = sqlQuery($sql);
return $result;
}
sqlQuery函数如下:
function sqlQuery($sql)
{
$products = array();
$link = dbConnect('localhost', 'root', '', 'umarrr');
$result = mysqli_query($link, $sql);
while ($row = mysqli_fetch_array($result))
{
$products[] = $row;
}
# return each row:
return $products;
#mysqli_close($link);
}
它们都已连接到数据库,并且一切正常。 但是,当我尝试遍历结果时:它仅返回一行:
$injury_table = injuryTable();
// make it more readable:
foreach ($injury_table as $table);
{
echo $table['pname'];
echo $table['name'];
echo $table['start_date'];
echo $table['end_date'];
}
我上面写的sql语句在mysql查询浏览器中可以完美地工作,所以有人知道这里可能是什么问题吗?
print_r($injury_table)
输出
Array ( [0] => Array ( [0] => 1 [id] => 1 [1] => 6 [injury_id] => 6 [2] => person [pname] => person [3] => wrist [name] => wrist [4] => 2008-11-21 [start_date] => 2008-11-21 [5] => 2010-11-11 [end_date] => 2010-11-11 ) [1] => Array ( [0] => 2 [id] => 2 [1] => 5 [injury_id] => 5 [2] => woman [pname] => woman [3] => neck [name] => neck [4] => 2009-11-12 [start_date] => 2009-11-12 [5] => 2010-09-09 [end_date] => 2010-09-09 ) [2] => Array ( [0] => 3 [id] => 3 [1] => 4 [injury_id] => 4 [2] => girl [pname] => girl [3] => groin [name] => groin [4] => 2010-11-27 [start_date] => 2010-11-27 [5] => 2010-12-01 [end_date] => 2010-12-01 ) [3] => Array ( [0] => 4 [id] => 4 [1] => 1 [injury_id] => 1 [2] => boy [pname] => boy [3] => achilles [name] => achilles [4] => 2010-02-01 [start_date] => 2010-02-01 [5] => 2010-03-23 [end_date] => 2010-03-23 ) [4] => Array ( [0] => 5 [id] => 5 [1] => 2 [injury_id] => 2 [2] => man [pname] => man [3] => toe [name] => toe [4] => 2010-01-01 [start_date] => 2010-01-01 [5] => 2010-02-02 [end_date] => 2010-02-02 ) )
要检查的一些事情:
mysqli_query()
的返回值。 您假设查询成功,并且有太多原因导致查询每次都不进行不检查。 如果失败,它将返回布尔值FALSE。 sqlQUery()
函数内的静态变量中并重复使用。在很多情况下,您都希望(或需要)多个可以同时使用,也可以每次使用全新的闪亮清洁手柄。 mysqli_num_rows()
来查看返回了多少PHP版本的查询? 为什么要为每个呼叫打开与数据库的新连接? 这是执行查询的非常低效的方式。 我会将数据库连接作为参数传递,或者由于您正在使用mysqli,因此只需将$link
作为参数传递。
至于为什么您的代码不起作用,我不知道,但是您可以尝试以下一些基本的错误报告:
$result = mysqli_query($link, $sql) or
trigger_error('Query Failed: ' . mysqli_error($link));
我还将MYSQL_ASSOC
添加MYSQL_ASSOC
函数中,因为您没有使用基于索引的数组,因此可以使脚本效率更高。
希望这可以帮助你。
真的很糟糕(也许是侮辱性的)答案...
foreach ($injury_table as $table);
{
echo $table['pname'];
echo $table['name'];
echo $table['start_date'];
echo $table['end_date'];
echo '<br/>'; // HTML new line
}
要么
foreach ($injury_table as $table);
{
echo $table['pname'];
echo $table['name'];
echo $table['start_date'];
echo $table['end_date'];
echo "\n"; // Console new line
}
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