这里的代码应该搜索数据库。但我得到我的表不存在的错误。另外我想问问为什么如果我推第二次提交按钮,它只是跳转到else
所以它回声choose at least.... and also all data from database.
谢谢!获取表不存在与PHP和MySQL
这里是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'];
}
}
}
这里是错误消息,当我到城市中的城市加名。
Table 'stanice_tab.station_tabwhere' doesn't exist
您确定要连接到正确的数据库吗? – D4V1D
是的,我是,因为它回声从station_tab的所有数据:) – R2D2
所以,当你得到“表不存在”的错误? – D4V1D