在此代码中，我想添加弹出框链接，但在内部显示一个错误<>.. 帮助我

Question

    if (strlen($string) > 500)
        {
            $stringCut = substr($string, 0, 500);
            $string = substr($stringCut, 0, strrpos($stringCut, ' ')).'... <a href="#" onclick="showAjaxModal('.base_url().'index.php?modal/popup/readmore/' . $row['categories_id'].')"> ...Read More </a> ' ; 
        }
    echo $string;
?>

未捕获的语法错误：无效或意外的令牌

Answer 1

像这样尝试：

$string = substr($string, 0, 500);
$string = substr($string, 0, strrpos($string, ' '));
$string .= '... <a href="#" onclick="showAjaxModal('' . base_url() . 'index.php?modal/popup/readmore/' . $row['categories_id'] . '')">... Read More</a>';

您缺少单引号，这些引号应将 url 包装在 showAjaxModal 函数中。

Answer 2

也许您忘记了 URL 的引号？

$string = substr($stringCut, 0, strrpos($stringCut, ' ')).'... <a href="#" onclick="showAjaxModal("'.base_url().'index.php?modal/popup/readmore/' . $row['categories_id'].'")"> ...Read More </a> ' ; 

Answer 3

这是另一个更易于管理的版本：

if (strlen($string) > 500) {
  $cut = substr($string, 0, 500);
  $cut = substr($cut, 0, strrpos($cut, ' '));
  $url = base_url() . "index.php?modal/popup/readmore/" . $row['categories_id'];
  $string = "$cut... <a href='#' onclick='showAjaxModal("$url")'> ...Read More </a>"; 
}
echo $string;