我正在尝试使用LEFT()函数从数据库中获取字段。我从描述字段中获取前20个字符。它获取并获得了结果,但有时候像这样得到了不完整的词 - 你过去的一些问题。
所以,我想要显示这样的完整单词 - 你过去的一些问题
我知道我取了前20个字,这就是为什么我拿不完整的字。但是,如果有任何解决方案可用于获取完整的单词,请建议我。
<?php
$query=mysqli_query($conn, "SELECT *, LEFT(descripation,20) AS descripation, LEFT(title,20) AS title FROM event WHERE status='Y'") or die("Could not retrieve image: " .mysqli_error($conn));
if (mysqli_num_rows($query) > 0){
while($row = mysqli_fetch_assoc($query)){
?>
<!--Event Thumb Start-->
<div class="col-md-4 col-sm-6">
<div class="msl-event">
<figure>
<img class="img-responsive" src="admin_panel/event/<?php echo $row['image'];?>">
<div class="theme-bg date-box">
<?php
echo $newDate = date("F d,Y", strtotime($row['date']));
?>
</div>
</figure>
<div class="text white_bg">
<h4 class="event-title"><a href="event-detail.php?id=<?php echo ($row['id']);?>"><?php echo $row['title'];?>...</a></h4>
<p><?php echo $row['descripation'];?>...</p>
</div>
</div>
</div>
<!--Event Thumb End-->
<?php
}
}
?>
答案 0 :(得分:1)
如何选择30个字符然后修剪字符串的最后一个字?
<?php
$query=mysqli_query($conn, "SELECT *, LEFT(description,30) AS description, LEFT(title,20) AS title FROM event WHERE status='Y'") or die("Could not retrieve image: " .mysqli_error($conn));
if (mysqli_num_rows($query) > 0){
while($row = mysqli_fetch_assoc($query)){
//trim description
$row['description'] = substr($row['description'], 0, strripos($row['description'], " ");
//trim description
?>
<!--Event Thumb Start-->
<div class="col-md-4 col-sm-6">
<div class="msl-event">
<figure>
<img class="img-responsive" src="admin_panel/event/<?php echo $row['image'];?>">
<div class="theme-bg date-box">
<?php
echo $newDate = date("F d,Y", strtotime($row['date']));
?>
</div>
</figure>
<div class="text white_bg">
<h4 class="event-title"><a href="event-detail.php?id=<?php echo ($row['id']);?>"><?php echo $row['title'];?>...</a></h4>
<p><?php echo $row['description'];?>...</p>
</div>
</div>
</div>
<!--Event Thumb End-->
<?php
}
}
?>
FYI描述拼写描述。