我正在从MYSQL进行简单搜索,它会从我的表中显示一个数据 " EMPLOYEE_ID"它已经有一个关于输出的链接我希望它指向具有相同id的另一个页面并从其被调用的id输出整个表
我已经有代码指向另一个页面但显示了我表格中的所有数据
我是php新手,我也是学生
形式:
<form action="searchemp.php?searchresult" method="post">
<p id="srchlbl">Search by Employee ID:</p>
<input id="srchtxt" type="text" name="employee_id" placeholder="Search for Employee..."><br>
<input id="srchbttn" name="search" type="submit" value="search">
<input name="viewall" type="submit" value="view all">
搜索
if(isset($_POST['search']))
{ if(isset($_GET['searchresult']))
{ if(preg_match("/^[ a-zA-Z]+/", $_POST['employee_id']))
{ $name = $_POST['employee_id'];
$sql = "SELECT * FROM employee WHERE employee_id LIKE '%" . $name . "%'";
$result=mysql_query($sql);
$row = mysql_fetch_array($result) or die($result."<br/><br/>".mysql_error());
$e_id = $row['employee_id'];
echo "<ul>\n";
echo "<b>search results</b>";
echo "<li>" . "<a href=\"singlesearchresult.php?action=employee_id\">".$e_id . "</a></li>\n";
echo "</ul>";
}
}
else{
echo "<p>Please enter a search query</p>";
}
}
singlesearchresult
<?php
$sql = 'SELECT employee_id, emp_fname, emp_lname, emp_address, emp_email, emp_contact FROM employee';
$retval = mysql_query( $sql);
if(! $retval ) {
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
echo
"<ul><li>EMPLOYEE ID : {$row['employee_id']} <br> ".
"EMPLOYEE NAME : {$row['emp_fname']} <br> ".
"EMPLOYEE LASTNAME : {$row['emp_lname']} <br> ".
"EMPLOYEE ADDRESS : {$row['emp_address']} <br> ".
"EMPLOYEE EMAIL : {$row['emp_email']} <br> ".
"EMPLOYEE CONACT : {$row['emp_contact']} <br> ".
"---------------------------------------------------------</li></ul>";
}
echo "<br><b>Fetched data successfully</b>\n";
?>
答案 0 :(得分:1)
您需要解析要使用的ID:
echo '<li><a href="singlesearchresult.php?id='.$e_id.'">'.$e_id.'</a></li>\n";
然后在singlesearchresult.php中使用查询中的$_GET['id']
$sql = 'SELECT employee_id, emp_fname, emp_lname, emp_address, emp_email, emp_contact FROM employee WHERE employee_id="'.$_GET['id'].'"';
NB:这只是一个疯狂不安全的例子,在查询中使用它之前,你必须先清理网址中传递的值
答案 1 :(得分:0)
在搜索
上使用此:
#LoadModule rewrite_module modules/mod_rewrite.so
在 singlesearchresult 页面上:
echo "<li><a href='singlesearchresult.php?action=".$e_id."'>".$row['emp_fname']."</a></li>\n";