传递PHP下一页变量

时间:2019-05-22 10:39:50

标签: javascript php mysql

我想将php变量传递给下一页。

我尝试了很多事情,但没有成功。 这是我尝试过的附件。

索引代码

<script>
<?php
$con = mysqli_connect($db_host,$db_user,$db_passwd,$db_name);

mysqli_set_charset($con, "utf8");

$result = mysqli_query($con, "select * from StoreTable");

$n = 1;
while($row = mysqli_fetch_array($result)){
  $name = $row['name'];
  ?>
  positions_1.push({ content:
    '<div class="wrap">' +
    ' <div><a href="next.php?name=<?= $name; ?>" target="_blank" class="link">next page</a></div>' +
    '</div>'
  });
  <?
$n++;
}
?>
</script>

下一个代码

<?php 
    echo $name;
?>

url已更改为变量,但显示空白屏幕。 (变量未传递到下一页。)

1 个答案:

答案 0 :(得分:0)

尝试一下-

<script>
<?php
$con = mysqli_connect($db_host,$db_user,$db_passwd,$db_name);

mysqli_set_charset($con, "utf8");

$result = mysqli_query($con, "select * from StoreTable");

$n = 1;
while($row = mysqli_fetch_array($result)){
  $name = $row['name'];
  ?>
  positions_1.push({ content:
    '<div class="wrap">' +
    ' <div><a href="next.php?name=<?php echo $name; ?>" target="_blank" class="link">next page</a></div>' +
    '</div>'
  });
  <?
$n++;
}
?>
</script>