Question

以下是我显示用户评论的查询。效果很好：

enter 
$query='SELECT
customers.cust_id,
customers.f_name,
customers.l_name,
user_comments.comment_txt,
clubs.name_club
from
customers
    inner join
user_comments
    on customers.cust_id = user_comments.cust_id
    inner join
clubs
    on user_comments.cust_id = clubs.cust_id
    ORDER BY RAND() LIMIT 1,1
    ';

$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$nameclub =$row['name_club'];
$comment =$row['comment_txt'];
echo '<div><img src="images/arrow.gif">&nbsp;<a   href="news.php?id='.$row['cust_id'].'">&nbsp;Name : '.$nameclub.'</a></div>';
echo '<div>'.$comment.'</div><br>';
echo '<div>'.$row['f_name'].' '.$row['l_name'].'</div>';
}
echo '<p style="text-align: left"><a href="newsarchive.php">&nbsp;Show all comments</ ></a></p> '; ?>

我现在想要显示用户评论的大约100个字符。所以我相应地更改了我的查询，但它不起作用：

enter $query='SELECT
customers.cust_id,
customers.f_name,
customers.l_name,
user_comments.comment_txt.substr(comment_txt,1,100) as comment_txt,
clubs.name_club
from
customers
    inner join
user_comments
    on customers.cust_id = user_comments.cust_id
    inner join
clubs
    on user_comments.cust_id = clubs.cust_id
    ORDER BY RAND() LIMIT 1,1
    ';

$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$nameclub =$row['name_club'];
$comment =$row['comment_txt'];
 echo '<div><img src="images/arrow.gif">&nbsp;<a href="news.php?id='.$row['cust_id'].'">&nbsp;Name : '.$nameclub.'</a></div>';
 echo '<div>'.$comment.'</div><br>';
 echo '<div>'.$row['f_name'].' '.$row['l_name'].'</div>';
 }
 echo '<p style="text-align: left"><a href="newsarchive.php">&nbsp;Show all comments</ ></a></p> ';   ?>

如何修复查询以获得评论的100个字符？

Answer 1

您未正确使用SUBSTR()。替换这个：

user_comments.comment_txt.substr(comment_txt,1,100) as comment_txt,

有了这个：

SUBSTR(user_comments.comment_txt,1,100) as comment_txt,

修改每条评论：如果文字继续超过100个字符，则添加...，请使用

CONCAT(SUBSTR(user_comments.comment_txt,1,100), IF(LEN(user_comments.comment_txt) > 100, '...', '')) as comment_txt,

Answer 2

试试这个

替换此

user_comments.comment_txt.substr(comment_txt,1,100) as comment_txt,

要

SUBSTRING(user_comments.comment_txt,1,100) as comment_txt,

Mysql function_substring

Answer 3

另一个解决方案是使用php函数将字符串剪切成多个单词：

<?php

    function cut_string($string_to_cut, $number_words, $append = "...") {

        $string = array();
        $string_return = "";
        $return_string = "";

        $string = explode(" ", $string_to_cut);

        if(count($string) < $number_words) {

            for($x=0; $x < count($string); $x++) {
                $string_return .= $string[$x] . " ";
            }

            $return_string = substr($string_return, 0, -1);
        } 
        else {

            for($x=0; $x < $number_words; $x++) {
                $string_return .= $string[$x] . " ";
            }

            $string = substr($string_return, 0, -1);
            $return_string = $string . $append;
        }
        return($return_string);
    }
?>

注意：这是基于字数的，而不是基于字符数的。

为什么我的查询不能用于检索子字符串？

3 个答案: