用php mysql填充选择框 [英] populate a select box with php mysql
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问题描述
我无法填写表单中的选择框，以显示来自护士表的护士的现有姓名。谁能告诉我我做错了什么？提前致谢！
以下是表格
< form method =post action =insert.php>
< br>
< tr>< td align =left>< strong>护士信息< / strong>< / td>
< / td>
< tr>
< td> nurse_name< / td>
< td>< select name =valuelist>
mysql_select_db（a& e，$ con）或die（'无法选择数据库。'）;
$ fetch_nurse_name = mysql_query（SELECT DISTINCT $ nurse_name FROM nurse）;
$ result = mysqli_query（$ con，$ query）或死（无效查询）;
while $（$ throw_nurse_name = mysqli_fetch_array（$ fetch_nurse_name））{
echo'< option value = \\'''。$ nurse_name ['nurse_name']。'>'。 $ throw_nurse_name [ 'nurse_name'] '< /选项>'。
}
回显< / select>;
mysqli_close（$ con）;
？>
< input type =submitvalue =提交>
< / form>< / body>< / html>
解决方案
试试这个：
< html>< head>< title>连接到数据库< / title>< / head>< body>
< font size =4>查询获取护士的姓氏< / font>
< br>< br>< font size =4>选择名称< / font>< br>< br>
< form action =insert.phpmethod =post>
< select name =valuelist>;
<？php
$ value = $ _ POST [valuelist];
$ con = mysql_connect（localhost，root，）或die（'Could not connect：'。mysql_error（））;
mysql_select_db（a& e，$ con）或die（'无法选择数据库。'）;
$ fetch_nurse_name = mysql_query（SELECT DISTINCT Forename FROM nurse）;
$ b $ while（$ throw_nurse_name = mysql_fetch_array（$ fetch_nurse_name））{
echo'< option value = \'''。$ throw_nurse_name [0]。'> 。$ throw_nurse_name [0]。 '< /选项>';
}
回显< / select>;
？>
< input type =submitvalue =提交>
< / form>< / body>< / html>
不要一起使用mysql和mysqli ....你应该使用mysqli或PDO，但不要混合;
PS：已编辑;）
Saludos。$ / b $ b
I'm having difficulty populating a select box within a form to display existing "forenames" of nurses from the "Nurses" table. Could anyone tell me what Im doing wrong? Thanks in advance!
Here is the form
<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>
The QUERY which should populate the nurse_forename:
<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>
解决方案
Try this:
<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>
Dont use mysql and mysqli together....you should use mysqli or PDO, but not a mix of both ;) PS: Edited ;)
Saludos.
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