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<html>
<head><title> Inpatient Search </title><head>
<body>
<h1> Inpatient Search </h1>
<?php
$wardnumber = $_POST['wardnumber'];
$conn = mysqli_connect("127.0.0.1","root","root","");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if ($q1 = mysqli_query($conn,"select s.Inpatient_id, s.Ward_number,c.Room_number,c.Floor_number,c.Class
from joseph.ward_stay_1501003f s, joseph.ward_1501003f c
where s.Ward_number = c.Ward_number and s.Ward_number='$wardnumber'")) {
print "<table border=1>";
print "<TR><TH>Inpatient id</TH><TH>Ward number</TH><TH>Room number</TH><TH>Floor number</TH><TH>Class</TH></TR>";
while ($r1 = mysqli_fetch_row($q1)) {
for ($k=0; $k<count($r1); $k++){
print htmlspecialchars($r1[$k]). " : ";
}
print "<BR>";
};
} else {
printf("Errormessage: %s\n", mysqli_error($conn));
}
mysqli_close($conn);
?>
</body>
</html>
我有一個有序的輸出。我如何解決它,使數據頂部的表頭?PHP奇怪的輸出
提示:繞過角色要求不能解決問題質量問題。 –
在while循環中,你爲什麼不回顯表格行?你從一張桌子開始,然後轉向一個列表。這不是一個PHP問題。這是關於您輸出的html的問題 –