Question

让我说我检查是否

$strig = "red-hot-chili-peppers-californication";

已存在于我的数据库中：

$query = dbquery("SELECT * FROM `videos` WHERE `slug` = '".$strig."';");
$checkvideo = dbrows($query);
if($checkvideo == 1){

// the code to be executed to rename $strig
// to "red-hot-chili-peppers-californication-2"
// it would be great to work even if $string is defined as 
// "red-hot-chili-peppers-californication-2"  and 
// rename $string to "red-hot-chili-peppers-californication-3"  and so on...

}

我想这样做可以为更友好的网址结构创建独特的slu ..

感谢。

Answer 1

我可以为您提供Codeigniter's increment_string()功能的来源：

/**
 * CodeIgniter String Helpers
 *
 * @package     CodeIgniter
 * @subpackage  Helpers
 * @category    Helpers
 * @author      ExpressionEngine Dev Team
 * @link        http://codeigniter.com/user_guide/helpers/string_helper.html
 */

/**
 * Add's _1 to a string or increment the ending number to allow _2, _3, etc
 *
 * @param   string  $str  required
 * @param   string  $separator  What should the duplicate number be appended with
 * @param   string  $first  Which number should be used for the first dupe increment
 * @return  string
 */
function increment_string($str, $separator = '_', $first = 1)
{
    preg_match('/(.+)'.$separator.'([0-9]+)$/', $str, $match);

    return isset($match[2]) ? $match[1].$separator.($match[2] + 1) : $str.$separator.$first;
}

通过向其附加数字或增加字符串来增加字符串   数。用于创建“副本”或文件或复制数据库   内容具有独特的标题或slu ..

用法示例：
echo increment_string('file', '_'); // "file_1"
echo increment_string('file', '-', 2); // "file-2"
echo increment_string('file-4'); // "file-5"

Answer 2

$str = "some-string-that-might-end-in-a-number";
$strLen = strlen($str);
//check the last character of the string for number
if(intval($str[$strLen-1])>0)
{
    //Now we replace the last number with the number+1
    $newNumber = intval($str[$strLen-1]) +1;
    $str = substr($str, 0, -1).$newNumber;
}
else
{
    //Now we append a number to the end;
    $str .= "-1";
}

当然这个限制是它只能得到最后一位数......如果这个数字是10怎么办？

$str = "some-string-that-might-end-in-a-number";
$strLen = strlen($str);

$numberOfDigits = 0;
for($i=$strLen-1; $i>0; $i--)
{
    if(intval($str[$i])==0)
    {
        $numberOfDigits = $strLen-($i-1);
        break;
    }
}

//Now lets do the digit modification
$newNumber = 0;
for($i=1; $i<=$numberOfDigits; $i++)
{
    $newNumber += intval($str[$strLen-$i])*((10*$i)-10));
}
if($newNumber == 0)
{ $newNumber = 1; }

$newStr = "-{$newNumber}";

//Now lets add the new string to the old one
$str = substr($str, 0, ($numberOfDigits*-1)).$newNumber;

如何检查一个字符串是否已经存在以及它是否在最后添加了+1？

2 个答案: