這裏的代碼應該搜索數據庫。但我得到我的表不存在的錯誤。另外我想問問爲什麼如果我推第二次提交按鈕,它只是跳轉到else
所以它回聲choose at least.... and also all data from database.
謝謝!獲取表不存在與PHP和MySQL
這裏是php
if (isset($_POST['submit'])) {
$query = 'SELECT * FROM station_tab';
if (!empty($_POST['station_name']) && !empty($_POST['city']) && !empty($_POST['zone']))
{
$query .= 'WHERE station_name' .mysql_real_escape_string($_POST['station_name']) . 'AND city' . mysql_real_escape_string($_POST['city']) . 'AND zone' . mysql_real_escape_string($_POST['zone']);
} elseif (!empty($_POST['station_name'])) {
$query .= 'WHERE station_name' . mysql_real_escape_string($_POST['station_name']);
} elseif (!empty($_POST['city'])) {
$query .= 'WHERE city' . mysql_real_escape_string($_POST['city']);
} elseif (!empty($_POST['zone'])) {
$query .= 'WHERE zone' . mysql_real_escape_string($_POST['zone']);
} else {
echo "Choose at least one option for search";
}
$result = mysql_query($query, $db) or die(mysql_error($db));
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)){
echo '<br/><em>' .$row['station_name'] . '</em>';
echo '<br/>city: '. $row['city'];
echo '<br/> zone: ' .$row['zone'];
echo '<br/> Long: ' .$row['lon'];
echo '<br/> Lat: ' . $row['lat'];
}
}
}
這裏是錯誤消息,當我到城市中的城市加名。
Table 'stanice_tab.station_tabwhere' doesn't exist
您確定要連接到正確的數據庫嗎? – D4V1D
是的,我是,因爲它回聲從station_tab的所有數據:) – R2D2
所以,當你得到「表不存在」的錯誤? – D4V1D