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php 检查 url 中是否存在变量并替换值

php 检查 url 中是否存在变量并替换值

PHP
达令说 2023-11-05 15:42:01
我写了一个函数,允许我进行分页。问题来了:当我单击按钮将页面更改为 URL 时,会添加许多相同的值,如下所示:时事通讯.php?action=sent-emails&page=2&page=1&page=4&page=5&page=7&page=5&page=1function pagination($sql_general,$page_name_pagination,$page_query_pagination) {  require $_SERVER['DOCUMENT_ROOT'] . 'it/intranet/includes/dbh.inc.php';  $sql1 =  $sql_general;  $sql1_query = mysqli_query($conn, $sql1);  $total = mysqli_num_rows($sql1_query);  $adjacents = 3;  $targetpage = $page_name_pagination; //your file name  if (!empty($page_query_pagination)) {    $targetpage .= "?" . $page_query_pagination . "&page=";  } else {    $targetpage .= "?page=";  }  $limit = 10; //how many items to show per page  $page = $_GET['page'];  if($page){  $start = ($page - 1) * $limit; //first item to display on this page  }else{  $start = 0;  }  /* Setup page vars for display. */  if ($page == 0) $page = 1; //if no page var is given, default to 1.  $prev = $page - 1; //previous page is current page - 1  $next = $page + 1; //next page is current page + 1  $lastpage = ceil($total/$limit); //lastpage.  $lpm1 = $lastpage - 1; //last page minus 1  $pagination = "";  if($lastpage > 1)  {  $pagination .= "<div class='pagination1'> <ul class='pagination'>";  if ($page > $counter+1) {  $pagination.= "<li class='page-item'><a class='page-link' href=\"$targetpage$prev\"><</a></li>";  }  if ($lastpage < 7 + ($adjacents * 2))  {  for ($counter = 1; $counter <= $lastpage; $counter++)  {  if ($counter == $page)  $pagination.= "<li class='page-item'><a class='page-link' href='#' >$counter</a></li>";  else  $pagination.= "<li class='page-item'><a class='page-link' href=\"$targetpage$counter\">$counter</a></li>";  }  }  elseif($lastpage > 5 + ($adjacents * 2)) //enough pages to hide some  {现在,如何检查“page=x”是否存在并替换该值?
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?
暮色呼如

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拆分、修改/添加和撰写:


parse_str($query_string,$list);

if (isset($list['name'])

{

     // name= exists, modify it

}

else

{

     // name= not exists

     $list['name'] = "new value";

}

$query_string = http_build_query($list);


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反对 回复 2023-11-05
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