用 php mysql 填充一个选择框
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populate a select box with php mysql
提问by user2075528
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!
我很难在表单中填充一个选择框来显示“护士”表中护士的现有“姓名”。谁能告诉我我做错了什么？提前致谢！
Here is the form
这是表格
<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>
The QUERY which should populate the nurse_forename:
应该填充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>
采纳答案by Hackerman
Try this:
尝试这个：
<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 ;)
不要同时使用 mysql 和 mysqli....你应该使用 mysqli 或 PDO，但不能混合使用 ;) PS：已编辑 ;)
Saludos.
萨卢多斯。
回答by Strawberry
Apologies if this duplicates other answers, Here's an answer using mysql_ syntax although you should of course be using mysqli_ or PDO for this...
抱歉，如果这与其他答案重复，这是使用 mysql_ 语法的答案，尽管您当然应该为此使用 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>
回答by MMReza
Check your MySQL table and column name that you using. Sometimes it does not work if you don't write those names exactly which in your MySQL table. suppose,
检查您使用的 MySQL 表和列名称。有时，如果您没有在 MySQL 表中准确写入这些名称，则它不起作用。认为，
$query = "SELECT nurse_name FROM nurse";
in above SQL if MySQL table name is 'NURSE' and column name is 'NURSE_NAME' then write exactly like this.
在上面的 SQL 中，如果 MySQL 表名是 'NURSE' 并且列名是 'NURSE_NAME' 然后完全像这样写。
$query = "SELECT NURSE_NAME FROM NURSE";
So, you look that sometime MySQL table, column name work in case sensitive.
所以，你看有时 MySQL 表，列名区分大小写。
