Question

我有一个sql查询女巫从sql数据库搜索学生名字它显示结果现在我想链接每个结果得到 http url www.tlss.edu.pk/result.php?r=1 如果 Sr_=1 我的SQL查询是

<?php
include 'form.html';
$r1=$_GET["r"];
$n1=$_GET["n"];
// Create connection
$con=mysqli_connect("localhost","chumspai_tlss","Tls121","chumspai_tlsResult");

// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
  echo $n1;
$result = mysqli_query($con,"SELECT * FROM nursery_blue_ WHERE students_names like '%$n1%'");

while($row = mysqli_fetch_array($result))
{

echo '<pre>';
    print_r ($row['sr_']. '.'. $row['students_names']);
    echo '</pre>';
  }
    ?>

Answer 1

可能需要尝试

while($row = mysqli_fetch_array($result))
{
     //print_r ($row['sr_']. '.'. $row['students_names']);
     if($row['sr_'] == 1) {
        echo "<a href='www.tlss.edu.pk/result.php?r=".$row['sr_']."'>"
              .$row['students_names']."</a>";
     }
 }

如果您想要所有结果，则可以在此处删除if条件。

Answer 2

在你的循环中添加一个链接，如果你有任何条件为show link添加条件

while($row = mysqli_fetch_array($result)) {
   echo "<a href='http://www.tlss.edu.pk/result.php?r=".$row['sr_']."'>".$row['students_names']."</a>";
 }

在PHP中链接SQL查询结果

2 个答案: