我正在尝试让我的搜索结果显示在表中的页面上,每行(行)有3个结果。我尝试了各种方法,修改了脚本很多次,但最终还是绕了一圈,回到了我原来的脚本。
最终,我希望将输出限制为每页50个,每行5个,但优先考虑的是正确显示输出。
任何帮助都是很棒的。
代码如下:
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
function searchmembers($search_term){
global $con;
$sql = mysqli_query($con, "SELECT * FROM `artist` WHERE `Band` LIKE '%$search_term%' OR `Genre` LIKE '%$search_term%' LIMIT 0, 30 ") or die (mysqli_error());
$num_of_row = mysqli_num_rows($sql);
if ($num_of_row > 0 ){
while($row = mysqli_fetch_array($sql))
{
$id = $row['Band'];
$Pic = $row['Pic'];
?>
<?php
echo"<table>";?>
<td><img src="<?php echo $row['Pic']; ?>" height="100" width="100" align="middle" /></td>
<?php
echo "<td><a href ='profile.php?Band=$id' style='color:white; text-decoration:none;'>". $row['Band']."";
echo"</table>";
}
}
else
{
echo "<font color='red' size='4' >No result found!</font>";
}
}
?>
这给出了这个结果:
发布于 2018-06-16 18:23:44
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
function searchmembers($search_term){
global $con;
$sql = mysqli_query($con, "SELECT * FROM `artist` WHERE `Band` LIKE '%$search_term%' OR `Genre` LIKE '%$search_term%' LIMIT 0, 30 ") or die (mysqli_error());
$num_of_row = mysqli_num_rows($sql);
if ($num_of_row > 0 ){
$i=0;
echo"<table>";
while($row = mysqli_fetch_array($sql))
{
$id = $row['Band'];
$Pic = $row['Pic'];
$i++;
?>
<?php if($i%3==1){ ?>
<tr>
<?php }?>
<td><img src="<?php echo $row['Pic']; ?>" height="100" width="100" align="middle" /></td>
<?php
echo "<td><a href ='profile.php?Band=$id' style='color:white; text-decoration:none;'>". $row['Band']."";
<?php if($i%3==1){ ?>
</tr>
<?php }?>
}
echo"</table>";
}
else
{
echo "<font color='red' size='4' >No result found!</font>";
}
发布于 2018-06-16 20:27:36
快到了,只需要在水平方向上连3个就行了?
https://stackoverflow.com/questions/50886807
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