我该如何将我的搜索结果变成这样的: http://i.stack.imgur.com/NfPGs.png
结果显示特定术语在单元格中的位置。
我目前有这个基本的搜索脚本:
$terms = explode("-", $SQuery);
$QuerySQL = "SELECT * FROM pages WHERE ";
foreach ($terms as $each){
$i++;
if ($i == 1)
$QuerySQL .= "Title LIKE '%$each%' OR Content LIKE '%$each%' OR Description LIKE '%$each%'";
else
$QuerySQL .= "OR Title LIKE '%$each%' OR Description LIKE '%$each%' OR Content LIKE '%$each%'";
}
$QueryNEW = mysql_query($QuerySQL);
WHILE($datarows_cat = mysql_fetch_array($QueryNEW)):
$title = $datarows_cat['Title'];
$Deleted = $datarows_cat['Deleted'];
$id = $datarows_cat['ID'];
if ($Deleted != "YES") {
echo "<a href='/{$id}'>{$title}</a><br/>";
}
最佳答案
您可以使用 PHP strpos查找 +/- 100 个字符:
$pos = strpos($mystring, $findme);
if ($pos === false) {
echo "The string '$findme' was not found in the string '$mystring'";
} else {
echo "The string '$findme' was found in the string '$mystring'";
echo " and exists at position $pos";
}
然后使用PHP substr获取 $pos+100
和 $pos-100
//For instance your substring could start at the position of the word found minus 100 characters and a length of 200
$result = substr($mystring, $pos-100, 200);
您可以使用 PHP str_replace格式化你需要的字符串:
$word_your_looking_for = "test";
$word_you_want_to_replace_it_with = "<b>test</b>"; //Make it bold
str_replace($word_your_looking_for, $word_you_want_to_replace_it_with, $your_string);
关于PHP 搜索获取文本之前/之后,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14944138/