[英]PHP - create one page that loads data from previous page
我有一個頁面,該頁面在用MySQL數據填充的表中顯示日期列表。
我在第1頁上顯示的字段是
firstname
lastname
email
在第2頁上,我想要記錄的完整信息。
我如何創建一個頁面,當用戶單擊firstname
作為<a href="">
鏈接時,它會自動加載信息,而不是為每個記錄創建單獨的頁面?
您可以嘗試將<site 1>?variable=value
附加到href上,然后通過PHP和?_GET['variable']
從第二個站點調用該<site 1>?variable=value
例如:
HTML
<a href="site2.php?firstname=foo">Click me!</a>
PHP
<?php echo $_GET['firstname']; ?>
您可以使用從mysql表獲取的記錄的一部分的id,然后將GET請求發送到第二頁,第二頁又將獲取表中的整個記錄...
/**
* ..Establish the connection to MySQL, Selects Database..
* ..Fetch few columns of all records..
*/
$result = mysql_query("SELECT id,firstname,lastname FROM records_table");
//Draw a table to hold the records
echo "<table>";
echo "<tr><th>Firstname</th><th>Lastname</th><th>Email</th><tr>";
while($row = mysql_fetch_assoc($result)){
echo "<tr>";
//Using the id of a record to create ahref link thats sends a get data to the second page
echo "<td><a href='second.php?id=".$row['id']."'>".$row['firstname']."</a></td>";
echo "<td>".$row['lastname']."</td>";
echo "<td>".$row['email']."</td>";
echo "</tr>";
}
echo "</table>";
$record_id = (int) $_GET['id'];
//Fetch the full details of this record...
$result = mysql_query("SELECT * FROM records_table WHERE id = ".$record_id."");
$row = mysql_fetch_assoc($result);
//List out the details..
echo "FIRSTNAME: ".$row['firstname'];
echo "LASTNAME: ".$row['lastname'];
echo "EMAIL: ".$row['email'];
//... as many column you may have...
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