PHP，而声明问题

Question

我有我的PHP while语句的问题。它是：

$row = mysqli_fetch_array($result); 
$time_in_12_hour_format = date("g:i a", strtotime($row['Time'])); 
while($row = mysqli_fetch_array($result)){ //Creates a loop to loop through results 
    echo "<tr><td>" . $row['Name'] . "</td><td>" . $row['Description'] . "</td><td>" . $row['Date'] . " at " . $time_in_12_hour_format . "</td><td>" . $row['Location'] . "</td>"; //$row['index'] the index here is a field name 

是的，我意识到我定义$行两次，但我不知道如何改变这一点。如果我从while语句中删除$ row = mysqli_fetch_array（$ result），页面将不会加载并返回错误。截至目前，页面加载但表格是空的，尽管我知道有一个记录在那里显示在表格中，然后我添加了军事时间转换器。任何帮助表示赞赏。

Answer 1

试试这个

while ($row = mysqli_fetch_array($result)) { //Creates a loop to loop through results 
    $time_in_12_hour_format = date("g:i a", strtotime($row['Time'])); 
    echo "<tr><td>" . $row['Name'] . "</td><td>" . $row['Description'] . "</td><td>" . $row['Date'] . " at " . $time_in_12_hour_format . "</td><td>" . $row['Location'] . "</td>"; 
} 

你叫mysqli_fetch_array它将从表中的一行，所以既然你说你的表中有一个记录，一个记录将在第一mysqli_fetch_array通话并最终第二返回每次致电mysqli_fetch_array返回NULL，因为表中没有更多记录要提取

Answer 2

尝试将$time_in_12_hour_format移动到while循环中。

while($row = mysqli_fetch_array($result)){ //Creates a loop to loop through results 
    $time_in_12_hour_format = date("g:i a", strtotime($row['Time'])); 
    echo "<tr><td>" . $row['Name'] . "</td><td>" . $row['Description'] . "</td><td>" . $row['Date'] . " at " . $time_in_12_hour_format . "</td><td>" . $row['Location'] . "</td>"; //$row['index'] the index here is a field name 
} 

这也将解决定义$行两次，这是首先导致问题的原因。因为你已经叫mysqli_fetch_array($result)你的while循环在第2行开始。

Answer 3

$time_in_12_hour_format应该为每一行设置吧？如果是，则尝试在while循环内移动它。您当前的代码只会开始显示第2行，第1行将被忽略。修改后的代码应该是这样的：

while($row = mysqli_fetch_array($result)){ //Creates a loop to loop through results 
    $time_in_12_hour_format = date("g:i a", strtotime($row['Time'])); 
    echo "<tr><td>" . $row['Name'] . "</td><td>" . $row['Description'] . "</td><td>" . $row['Date'] . " at " . $time_in_12_hour_format . "</td><td>" . $row['Location'] . "</td>"; //$row['index'] the index here is a field name 
}