好的。此代碼返回爲foreach提供的無效參數的語法錯誤()語法錯誤我找不到
我已經經歷了這個代碼很多,不能爲我的生活發現錯誤。
我還在學習和嘗試的東西,所以請溫柔。
<?php
include ('c1.php');
if ($_COOKIE["auth"] == "1") {
$display_block = "<p>You are an authorized user.</p>";
} else {
header("Location: userlogin.html");
exit;
}
doDB();
$display_block = "<h1>Results</h1>";
if (isset($_POST['search']) && !empty($_POST['search'])) {
foreach ($_POST['search'] as $key => $value) {
if ($value == 1)
$search[] = "$key";
$searchstring = implode(' AND ', $search);
$post_map = array(
'postcode' => 'candididate_contact_details.postcode'
);
if (isset($_POST['postcode']) && !empty($_POST['postcode'])) {
foreach ($_POST['postcode'] as $key => $value) {
if (array_key_exists($key, $post_map))
$search[] = $post_map[$key] . '=' . mysql_real_escape_string($value);
echo $searchstring;
$query = "SELECT candidate_id.master_id, candidate_contact_details.first_name, candidate_contact_details.last_name, candidate_contact_details.home_phone, candidate_contact_details.work_phone, candidate_contact_details.mobile_phone, candidate_contact_details.email FROM candidate_id, candidate_contact_details, qualifications, security_experience, previous_career WHERE qualifications.active = 'finished' and candidate_id.master_id = candidate_contact_details.master_id and candidate_id.master_id = qualifications.master_id and candidate_id.master_id = security_experience.master_id and candidate_id.master_id = previous_career.master_id and $searchstring";
$query_res = mysqli_query($mysqli, $query)
or die(mysqli_error($mysqli));
// $search = mysqli_query($mysqli, $query)or die(mysqli_error($mysqli));
{
$display_block .= "
<table width=\"98%\" cellspacing=\"2\" border=\"1\">
<tr>
<th>Registration Number</th>
<th>First Name</th>
<th>Last Name</th>
<th>Home Number</th>
<th>Work Number</th>
<th>Mobile Number</th>
<th>E-Mail</th>
</tr>";
while ($result = mysqli_fetch_array($query_res)) {
$regnum = $result['master_id'];
$first_name = $result['first_name'];
$last_name = $result['last_name'];
$home_phone = $result['home_phone'];
$work_phone = $result['work_phone'];
$mobile_phone = $result['mobile_phone'];
$email = $result['email'];
$display_block .= "
<tr>
<td align=\"center\">$regnum <br></td>
<td align=\"center\">$first_name <br></td>
<td align=\"center\">$last_name <br></td>
<td align=\"center\">$home_phone <br></td>
<td align=\"center\">$work_phone <br></td>
<td align=\"center\">$mobile_phone <br></td>
<td align=\"center\">$email <br></td>
</tr>";
}
$display_block .= "</table>";
}
}
}
}
}
?>
<html>
<head>
<title> Display results</title>
</head>
<body>
<?php echo $display_block; ?>
</body>
</html>
哪一行*是語法錯誤? –
對不起忘了。 foreach($ _POST ['postcode'] as $ key => $ value){ –
$ _POST ['postcode']上var_dump的輸出是什麼?它是一個數組嗎? –