Question

所以我有一个名为query_logs的表，其中包含三列--ID，query，date。这是我选择查询和日期的代码：

$qlogs = mysql_query('select query, date from fcms.query_logs order by id desc');
$query_logs = mysql_fetch_array($qlogs);

我正在尝试显示这些列中的所有信息：

for($counter = 1; $counter <= $count[0]; $counter++) {
echo $query_logs['date'].' - [ '.$query_logs['query'].' ] <br />'; 
}

但它只是一遍又一遍地重复同样的事情，直到$ counter = $ count ..例如：

2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]
2014-11-24 12:55:59 - [ select * from world.version ]

问题出在这里：

echo $query_logs['date'].' - [ '.$query_logs['query'].' ] <br />';

我不知道如何使其有效; /

完整代码：

<link rel="stylesheet" href="style.css" type="text/css" />
<form name="form" action="" method="get">
  <label for="query">Query</label><textarea name="query" cols="40" rows="5"></textarea><br />
  <label for="pass">Password</label><input type="password" name="pass" id="pass" /><br />
  <input type="submit" value="Enter"/>
</form>
<?php
include('query_configuration_xf1hfa2xaz.php');
$query = $_GET['query'];
$password = $_GET['pass'];
$date = date('Y-m-d H:i:s');
$qlogs = mysql_query('select query, date from fcms.query_logs order by id desc');
$query_logs = mysql_fetch_array($qlogs);
$countq = mysql_query('select count(id) from fcms.query_logs');
$count = mysql_fetch_row($countq);
if($password != '') {
if($password == $security) {
$runquery = mysql_query($query);
$result = mysql_fetch_array($runquery);
if($runquery == true) 
{
    echo '<br /> The query was successfully executed! <br />';
    mysql_query('insert into fcms.query_logs (query, date) values ("'.$query.'", "'.$date.'") ');
} 
else 
{ 
    echo '<br />There\'s some error with your query! Check it again..<br />'.mysql_errno()." : "
         .mysql_error();
}
}
else
{
    echo 'The password is wrong!';
}
echo $result[0].'<br />';
}
for($counter = 1; $counter <= $count[0]; $counter++) {
echo $query_logs['date'].' - [ '.$query_logs['query'].' ] <br />'; 
}
?>

Answer 1

您需要遍历mysql_fetch_array()

$counter，这里没有任何意义，所以，请将其删除。

$qlogs = mysql_query('select query, date from fcms.query_logs order by id desc');
while ($query_logs = mysql_fetch_array($qlogs)) {
  echo $query_logs['date'].' - [ '.$query_logs['query'].' ] <br />'; 
}

Answer 2

你似乎错误地解释了 mysql_fetch_array（）的作用。它只为表的一行返回一个值数组，而不是一个行数组，您可以通过这些行与计数器进行迭代（另外，您不要在循环中使用计数器）。因此，您必须再次为要阅读的每一行调用它。使用while循环，您可以轻松地遍历整个表，因为 mysql_fetch_array（）只有在读取了所有行后才会返回 false 。

while ($row = mysql_fetch_array($qlogs)) {
   echo $row['date'].' - [ '.$row['query'].' ] <br />'; 
}

从中选择两列并显示它

2 个答案: