Question

我对PHP还是很陌生，所以如果我要问一些明显的问题，请原谅！

我试图从下拉控件中获取用户选择的值，并将其存储在数据库的表中。问题在于，下拉列表中的值本身是从同一数据库中的不同表中查询的。我可以使用下拉列表来正确填充，但是随后我很难使表单提交过程正确地引用该值，然后将其存储在表中。我希望有人可以帮助我。

我在下面提供了我正在努力处理的代码，这些代码是从网络上的许多不同来源中剔除的，因此，如果从专家的角度来看它完全错误，请不要感到惊讶：）< / p>

这是显示表单和自动填充的下拉菜单的代码

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$conn = sqlsrv_connect( $serverName, $connectionInfo);


if( $conn ) {
     echo "Connection established.<br /><br>";
}else{
     echo "Connection could not be established.<br />";
     die( print_r( sqlsrv_errors(), true));
}

$sql = "Select ProductName from ProductTypes";
$dropdown = sqlsrv_query($conn, $sql);


echo "Please select the Product Type ";
echo "<Select Product Type='ProductName'>";
while ($row = sqlsrv_fetch_array($dropdown)) {
    echo "<option value='" . $row['ProductName'] . "'>" . $row['ProductName'] ."</option>";
}
echo "</select>";
?>

<form action="insertmssql.php" method="post">

<br>
ProjectID: <input type="text" name="ProjectID" /><br><br>

TestVal1: <input type="number" step="0.01" name="TestVal1" /><br><br>

TestVal2: <input type="number" step="0.01" name="TestVal2" /><br><br>

Result: <input type="text" name="Result" /><br><br>

<input type="submit" />

</form>

</body>
</html>

¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬¬ ¬¬¬¬p

这是执行提交的代码

$conn = sqlsrv_connect( $serverName, $connectionInfo);


if( $conn ) {
     echo "Connection established.<br />";
}else{
     echo "Connection could not be established.<br />";
     die( print_r( sqlsrv_errors(), true));
}


$ProductName = $_REQUEST['dropdown'];
$ProjectID = $_REQUEST['ProjectID'];
$TestVal1 = $_REQUEST['TestVal1'];
$TestVal2 = $_REQUEST['TestVal2'];
$Result = $_REQUEST['Result'];



// Attempt insert query execution
$sql = "INSERT INTO testdata (ProductName, ProjectID, TestVal1, TestVal2, Result) VALUES ('$ProductName', '$ProjectID', '$TestVal1', '$TestVal2', '$Result')";
if(sqlsrv_query($conn, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . sqlsrv_errors($conn);
}

// Close connection
sqlsrv_close($conn);

PHP-从数据库生成的下拉列表中获取价值并插入到另一个表中

0 个答案: