如何在 table 中显示 mysql 多行/mysql_fetch_array 结果？

Question

我试图在 table 中显示 mysql_fetch_array() 结果。

我想展示 guests name、their country 和他们 agreed time 在同一天旅行的 。

以下代码工作正常。代码获取行值并打印它。

 $select_guests = mysql_query('SELECT name FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_guests, MYSQL_ASSOC)) { //visitor / guest loop starts here
    echo $row['name'].'<br/>';
}
$select_country = mysql_query('SELECT country FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_country, MYSQL_ASSOC)) { //country of visitor / guest loop starts here
    echo $row['country'].'<br/>';
}
$select_agreed_time = mysql_query('SELECT agreed_time FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_agreed_time, MYSQL_ASSOC)) { //visitor / guest agreed time loop starts here
    echo $row['agreed_time'].'<br/>';
}

如果同一天有 5 位客人，当我执行上述代码时，我会得到他们所有 names 一个低于另一个的。我也到了 countries 和 agreed time。

现在我想在 HTML table 中显示这些结果。我尝试了几行代码但没有任何效果。

我的HTMLtable应该如下：

<table class="table-fill">
    <thead>
        <tr>
            <th class="text-left">Name</th>
            <th class="text-left">From</th>
            <th class="text-left">Agreed Time</th>
        </tr>
    </thead>
    <tbody class="table-hover">
        <tr>
            <td class="text-left">Name 1</td>
            <td class="text-left">Country 1</td>
            <td class="text-left">Ag Time 1</td>
        </tr>
        <tr>
            <td class="text-left">Name 2</td>
            <td class="text-left">Country 2</td>
            <td class="text-left">Ag Time 2</td>
        </tr>
        <tr>
            <td class="text-left">Name 3</td>
            <td class="text-left">Country 3</td>
            <td class="text-left">Ag Time 3</td>
        </tr>
        <tr>
            <td class="text-left">Name 4</td>
            <td class="text-left">Country 4</td>
            <td class="text-left">Ag Time 4</td>
        </tr>
        <tr>
            <td class="text-left">Name 5</td>
            <td class="text-left">Country 5</td>
            <td class="text-left">Ag Time 5</td>
        </tr>
    </tbody>
</table>

如何根据我的 mysql_fetch_array() 创建 table td ？上面的 table 结构是 5 guests 由 mysql_fetch_array()

找到或产生的

Answer 1

首先，我认为您的解决方案不需要 3 个不同的查询..

    <table class="table-fill">
            <thead>
                <tr>
                    <th class="text-left">Name</th>
                    <th class="text-left">From</th>
                    <th class="text-left">Agreed Time</th>
                </tr>
            </thead>
        <?php 
        $result = mysql_query('SELECT name,country,agreed_time FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); 
        while($row = mysql_fetch_array($result, MYSQL_ASSOC)) 
        { 
        ?>
          <tr>
               <td>
                   <?php    echo $row['name']; ?>
               </td>
               <td>
                   <?php     echo $row['country'];?>
               </td>
                <td>
                   <?php     echo $row['agreed_time']; ?>
               </td>    
    </tr>
    <?php
    }
?>
</table>

Answer 2

$select_all = mysql_query('SELECT * FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting all details for the same date
while($row = mysql_fetch_array($select_all , MYSQL_ASSOC)) {//loop starts here
    <tr>
           <td>
               <?php    echo $row['name']; ?>
           </td>
           <td>
               <?php     echo $row['country'];?>
           </td>
            <td>
               <?php     echo $row['agreed_time']; ?>
           </td>    
  </tr>

}

Answer 3

首先，你应该使用mysqli。

其次，不应该向数据库发送如此多的查询以获取一次查询中可以获得的信息；

$select_guests = $mysqli->query('SELECT * FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error());

接下来，您想fetch the number of rows。

$rows = $mysqli

您还应该查看 PHP for function;

for ($i = 1; $i < $rows; $i++) {
    $thisRow = $select_guests->fetch_row()
    echo
'        <tr>
        <td class="text-left">'.$select_guests['name'].'</td>
        <td class="text-left">'.$select_guests['country'].'</td>
        <td class="text-left">'.$select_guests['time'].'</td>
    </tr>
    '; //This last line is to insert a line break and indent (for tidy HTML)
}

试一试，希望对您有所帮助。不过我还没有完全为你解决这个问题，为了转换到 mysqli，你需要做一些小的改变，你可以在我发给你的 mysqli link 中找到这些改变。好处是值得的。

如何在 table 中显示 mysql 多行/mysql_fetch_array 结果？

How to show mysql multi row / mysql_fetch_array results in a table?

php

mysql

arrays

html-table

while-loop