[英]Displaying query results after submitting a PHP form
我目前正在使用php和mysql進行學校項目。 我創建了一個帶有三個下拉框的表單,用戶可以在其中選擇要查找的數據類型。 但是,提交表單后,我很難顯示結果。 這是我當前的代碼:
<?php
require_once 'connection.php';
?>
<form action="stats.php" method ="post">
<input type="hidden" name="submitted" value="true" />
<fieldset>
<legend>
Specify Date, Month, and County
</legend>
<p>
<label for="year">
Please select a year
</label>
<select name= 'year'>
<?php
$query = "select distinct year from unemployed";
$result = $conn->query($query);
while($row = $result->fetch_object()) {
echo "<option value='".$row->year."'>".$row->year."</option>";
}
?>
</select>
</p>
<p>
<label for="month">
Please select a month
<label>
<select name= 'month'>
<?php
$query = "select distinct month from unemployed";
$result = $conn->query($query);
while($row = $result->fetch_object()) {
echo "<option value='".$row->month."'>".$row->month."</option>";
}
?>
</select>
</p>
<p>
<label for="location">
Please specify a location
</label>
<select name='select'>
<?php
$query = "select * from unemployed";
$result = $conn->query($query);
while ($finfo = $result->fetch_field()) {
echo "<option value='".$finfo->name."'>".$finfo->name."</option>";
}
?>
</select>
</p>
<input type ="submit" />
</fieldset>
</form>
<?php
if (isset($_POST['submitted'])) {
include('connection.php');
$gYear = $_POST["year"];
$gMonth = $_POST["month"];
$gSelect = $_POST["select"];
$query = "select $gSelect from unemployed where year='$gYear' and month='$gMonth'";
$result = $conn->query($query) or die('error getting data');
echo"<table>";
echo "<tr><th>Year</th><th>Time</th><th>$gSelect</th></tr>";
while ($row = $result->fetch_object()){
echo "<tr><td>";
echo $row['Year'];
echo "</td><td>";
echo $row['Month'];
echo "</td><td>";
echo $row['$gSelect'];
echo "</td></tr>";
}
echo "</table";
} // end of main if statement
?>
我幾乎可以確定我的問題出在我的while陳述中。 我可以顯示表列的標題(年,月,$ gSelect),但是查詢結果卻無法顯示。
我努力了:
while ($row = $result->fetch_object())
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
這些都不適合我。 我看了php.net以獲得指導。 我仍然對做什么感到困惑。 如果有人可以幫助我,我將非常感激。
始終檢查您的退貨:
if( ! $result = $conn->query($query) ) {
die('Error: ' . $conn->error());
} else {
while($row = $result->fetch_object()) {
echo "<option value='".$row->year."'>".$row->year."</option>";
}
}
還放了error_reporting(E_ALL);
在您開發腳本的頂部時,它也會有很大幫助。
您應該真正考慮將變量作為參數傳遞給查詢,而不是將它們作為變量直接注入到查詢中。 這可能導致sql注入攻擊。
另外,這是一個如何使用PDO和mysql編寫查詢的快速示例:
//Simple Query
$dbh = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'username', 'password');
//Useful during development.
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//mysql can have prepares depending on the version: http://stackoverflow.com/questions/10113562/pdo-mysql-use-pdoattr-emulate-prepares-or-not
$dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$sth = $dbh->query("SELECT * FROM table");
var_dump($sth->fetchAll(PDO::FETCH_ASSOC));
//Now with pass params and a prepared statement:
$query = "SELECT * FROM table WHERE someCol = ?";
$sth = $dbh->prepare($query);
$sth->bindValue(1,"SomeValue");
$sth->execute();
$results = $sth->fetchAll(PDO::FETCH_ASSOC));
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