用php mysql填充選擇框
[英]populate a select box with php mysql
問題描述
我在填充表格中的選擇框以顯示“護士”表中護士的現有“姓氏”時遇到困難。 誰能告訴我我在做什么錯？ 提前致謝！
這是表格
<form method="post" action="insert.php">
<br>
<tr><td align="left"><strong>Nurse Information</strong></td>
</td>
<tr>
<td>nurse_name</td>
<td><select name="valuelist">
<option value="valuelist" name="nurse_name" value='<?php echo $nurse_name; ?>'></option>
</select></td>
<tr>
應該填充Nurse_forename的QUERY：
<html><head><title>Connect to Database</title></head><body>
<font size="4">Query gets Forename of nurse</font>
<br><br><font size="4">Choose a name</font><br><br>
<form action="insert.php" method="post">
<select name="valuelist">;
<?php
$value=$_POST ["valuelist"];
$con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
mysql_select_db("a&e", $con) or die('Could not select database.');
$fetch_nurse_name = mysql_query("SELECT DISTINCT $nurse_name FROM nurse");
$result = mysqli_query($con, $query) or die("Invalid query");
while($throw_nurse_name = mysqli_fetch_array($fetch_nurse_name)) {
echo '<option value=\"'.$nurse_name['nurse_name'].'">'.$throw_nurse_name['nurse_name'].'</option>';
}
echo "</select>";
mysqli_close($con);
?>
<input type="submit" value="Submit">
</form></body></html>
3 個解決方案
解決方案1
1 已采納 2013-02-15 13:17:10
嘗試這個：
<html><head><title>Connect to Database</title></head><body>
<font size="4">Query gets Forename of nurse</font>
<br><br><font size="4">Choose a name</font><br><br>
<form action="insert.php" method="post">
<select name="valuelist">;
<?php
$value=$_POST ["valuelist"];
$con = mysql_connect("localhost","root","") or die('Could not connect:'.mysql_error());
mysql_select_db("a&e", $con) or die('Could not select database.');
$fetch_nurse_name = mysql_query("SELECT DISTINCT Forename FROM nurse");
while($throw_nurse_name = mysql_fetch_array($fetch_nurse_name)) {
echo '<option value=\"'.$throw_nurse_name[0].'">'.$throw_nurse_name[0].'</option>';
}
echo "</select>";
?>
<input type="submit" value="Submit">
</form></body></html>
不要同時使用mysql和mysqli ....您應該同時使用mysqli或PDO，但不能同時使用它們；）PS：已編輯;）
Saludos。
解決方案2
0 2013-02-15 13:23:15
抱歉，如果這與其他答案重復，這是使用mysql_語法的答案，盡管您當然應該為此使用mysqli_或PDO ...
<form action="insert.php" method="post">
<select name="valuelist">;
<?php
//path to connection statements
include('path/to/connection/stateme.nts');
//fetch nurse name
$query = "SELECT nurse_name FROM nurse;";
$result = mysql_query($query) or die(mysql_error()); //note: use mysql_error() for development only
//print results
while($row = mysql_fetch_assoc($result)) {
echo '<option value=\"'.$row['nurse_name'].'">'.$row['nurse_name'].'</option>';
}
echo "</select>";
?>
<input type="submit" value="Submit">
</form>
解決方案3
0 2013-09-24 16:30:28
檢查您使用的MySQL表和列名。 有時，如果您沒有在MySQL表中寫出准確的名稱，那將是行不通的。 假設，
$query = "SELECT nurse_name FROM nurse";
在上面的SQL中，如果MySQL表名是“ NURSE”，列名是“ NURSE_NAME”，則應完全這樣寫。
$query = "SELECT NURSE_NAME FROM NURSE";
因此，您看起來有時候MySQL表，列名可以區分大小寫。
根據不同選擇框（php / ajax）的選擇填充選擇框
[英]Populate select box depending on choice of different select box (php / ajax)
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