0

我想使用 PHP 从数据库中填充两列 HTML 布局。返回的记录将按日期降序排列，我想填充列，以便第一条记录进入第一列，第二条记录进入第二列，然后返回第一列，依此类推。

我正在研究我应该获取 MySQL 结果并运行将其拆分为两个数组的理论，方法是将第一条记录放在第一个数组中，将第二条记录放在第二个数组中，依此类推，然后使用这些数组输出到列。

列定义如下；

<div id="leftcol">

</div>

<div id="rightcol">

</div>

编辑

while($row = mysql_fetch_array($result))
{
    $lcol = $lcol + 1;
    $vis = 0;
    $uri = substr($row[1], 0, strpos($row[1], "&"));
    $sql = "SELECT * FROM `readart` WHERE `url`=\"".$uri."\"";
    $result2 = mysql_query($sql);

    while($row2 = mysql_fetch_assoc($result2))
    {
        $vis = $row2['visits'];
    }

    $tit = myTruncate2($row[4], 91);
    echo '<div class="newitem">';
    echo '<img style="float:left;margin:6px;margin-right:20px;" src="newslogo/'.$row[2].'.png" width="40px" height="40px" />';
    echo '<span id="header">'.$tit.'</span><br>';
    //echo $row[6];
    echo '<span id="date">'.date("D, j F g:i A", $row[6]);
    if($vis > 0){echo ' - Viewed '.$vis.' times';}
    echo '</span><hr>';
    echo '<p id="textbody">';
    echo $row[5];
    echo '<br><br><a href="recordarticles.php?url='.$row[1].'" target="_blank">Read More</a>';
    echo '</p><br></div>';
}
4

2 回答 2

0

做这个

 $rightCol = "";
 $leftCol = "";
 $c = 0;
 foreach($data as $box){
       if ($c%2){
            $leftCol.= $box;
       } else {
            $rightCol.= $box;
       }
       $c++;
 }

然后为您的 html 执行此操作

<div id="leftcol">
<?= $leftCol ?>
</div>

<div id="rightcol">
<?= $rightCol ?>
</div>
于 2013-06-10T05:05:17.707 回答
0 
<?php
$row = array();
$rowCount = $mysql_row_count($result);
while ($row[] = mysql_fetch_array($result));

echo "<div idi\"leftcol\">";
    for ($i = 0; $i < $rowCount; $i+=2)
        echo $row[$i]; //Change this to whatever you want to echo or however you want to echo it
echo "</div>";

echo "<div idi\"rightcol\">";
    for ($i = 1; $i < $rowCount; $i+=2)
        echo $row[$i]; //Change this to whatever you want to echo or however you want to echo it
echo "</div>";
?>

编辑

这是我想出的你的代码......我不确定它是否会立即工作，因为我没有测试过它。但我希望它能让你知道现在该怎么做。

<?php
function echoData($right, $row, $rowCount)
{
    for ($i = $right; $i < $rowCount; $i += 2)
    {
        $lcol = $lcol + 1;
        $vis = 0;
        $uri = substr($row[$i][1], 0, strpos($row[$i][1], "&"));
        $sql = "SELECT * FROM `readart` WHERE `url`=\"".$uri."\"";
        $result2 = mysql_query($sql);

        while($row2 = mysql_fetch_assoc($result2))
        {
            $vis = $row2['visits'];
        }

        $tit = myTruncate2($row[$i][4], 91);
        echo '<div class="newitem">';
        echo '<img style="float:left;margin:6px;margin-right:20px;" src="newslogo/'.$row[$i][2].'.png" width="40px" height="40px" />';
        echo '<span id="header">'.$tit.'</span><br>';
        //echo $row[6];
        echo '<span id="date">'.date("D, j F g:i A", $row[$i][6]);
        if($vis > 0){echo ' - Viewed '.$vis.' times';}
        echo '</span><hr>';
        echo '<p id="textbody">';
        echo $row[$i][5];
        echo '<br><br><a href="recordarticles.php?url='.$row[$i][1].'" target="_blank">Read More</a>';
        echo '</p><br></div>';
    }
}

**Here you should put the query code**
$sql = "SELECT e.t.c e.t.c";
$result = mysql_query($sql);


$rowCount = mysql_num_rows(result);
$row = array();
while($row[] = mysql_fetch_array($result));

echo "<div id='leftside'>";
echoData(0, $row, $rowCount);
echo "</div>";

echo "<div id='rightside'>";
echoData(1, $row, $rowCount);
echo "</div>";
于 2013-06-10T05:10:35.053 回答