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php - 获取表不存在于 php 和 mysql

转载 作者:行者123 更新时间:2023-11-28 23:49:55 26 4
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这里的这段代码应该搜索数据库。但我收到错误消息说我的表不存在。我还想问为什么如果我第二次按下提交按钮它只是跳转到 else 所以它回显 choose at least.... 以及数据库中的所有数据。谢谢!

这是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'];
}
}
}

这是我将城市名称添加到城市时的错误消息。

Table 'stanice_tab.station_tabwhere' doesn't exist

最佳答案

这是您更正后的代码:

   $query = 'SELECT * FROM station_tab '; // note the space at the end
if (!empty($_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']).'"'; // note the = signs and the space before each AND
} elseif (!empty($_POST['station_name'])) {
$query .= ' WHERE station_name = "' . mysql_real_escape_string($_POST['station_name']).'"'; // note the = sign and the space at the beginning
} elseif (!empty($_POST['city'])) {
$query .= ' WHERE city = "' . mysql_real_escape_string($_POST['city']).'"'; // note the = sign and the space at the beginning
} elseif (!empty($_POST['zone'])) {
$query .= ' WHERE zone = "' . mysql_real_escape_string($_POST['zone']).'"'; // note the = sign and the space at the beginning
} else {
echo "Choose at least one option for search";
}

养成echo您的$query 变量的习惯,这样串联就不会添加任何打字错误。

关于php - 获取表不存在于 php 和 mysql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32801654/

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