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嗨,我想弄清楚这是否会在某种意义上起作用。现在当然不是。我现在抛出错误,但我也没有得到任何使用 $nextLink 变量输出的东西。

这是完整的,所以我们很清楚它们是如何结合在一起的:

<?php
// Check to see the URL variable is set and that it exists in the database
if (isset($_GET['id'])) {
// Connect to the MySQL database
include "includes/db_conx.php";
$id = intval($_GET['id']);// filter everything but numbers
// Use this var to check to see if this ID exists, if yes then get the product
// details, if no then exit this script and give message why


$sql = "UPDATE content SET views=views+1 WHERE ID=$id";
$update = mysqli_query($db_conx,$sql);

$sql = "SELECT * FROM content WHERE id=$id LIMIT 1";
$result = mysqli_query($db_conx,$sql);
$productCount = mysqli_num_rows($result);
//


if ($productCount > 0) {
// get all the product details
while($row = mysqli_fetch_array($result)){
$id = $row["id"];
$article_title = $row["article_title"];
$category = $row["category"];
$readmore = $row["readmore"];
$author = $row["author"];
$date_added = $row["date_added"];
$article = $row["article"];
$newDate = substr($date_added, 0, 10); 
}
} else {
echo "That item does not exist.";
exit();
}

} else {
echo "Data to render this page is missing.";
exit();
}
$sqltwo = "SELECT * FROM content WHERE id =(select min(id) from content where id > '$id') LIMIT 1";
$next = mysqli_query($db_conx,$sqltwo);
if($next > 0){
while($row = mysqli_fetch_array($next)){
$id = $row["id"];
$nextLink = '$id';
}
}

?>

这是我正在处理的片段。

$sqltwo = "SELECT * FROM content WHERE id =(select min(id) from content where id > '$id') LIMIT 1";
$next = mysqli_query($db_conx,$sqltwo);
if($next > 0){
while($row = mysqli_fetch_array($next)){
$id = $row["id"];
$nextLink = '$id';
}
}

我试图通过在当前 id 之后抓取一堆并选择下一个来为 HTML 提供 $nextLink 来选择下一行。

这是我 $next 的 vardump

object(mysqli_result)#3 (5) {
  ["current_field"]=>
  int(0)
  ["field_count"]=>
  int(10)
  ["lengths"]=>
  NULL
  ["num_rows"]=>
  int(0)
  ["type"]=>
  int(0)
}
4

1 回答 1

0

您没有得到所需的$nextLink变量输出的原因是您已将其括在单引号'中。

双引号内的变量"由解析器评估。双引号内的变量'不会被解析器评估,而是被视为文字字符串。

而不是这个

$nextLink = '$id';

用这个

$nextLink = $id;甚至可以这样做(但不建议)$nextLink = "$id";

您将获得所需的输出$nextLink。希望有帮助:)

于 2013-08-10T02:51:28.573 回答