具有多个输入的 PHP 搜索表单

Question

我有一个 PHP 脚本，我正在尝试使用它从数据库生成搜索结果，其中包含多个搜索文本输入和一个提交。三个输入是术语、关键字和位置。我已经完成了搜索，但我发现验证逻辑有点困难。我将查询分为三个部分，以便当其中一个输入为空时，它可以免除将查询的那部分添加到完整查询中。但是由于 OR 运算符，当第一个输入为空时，其余的都失败了。

第二只眼睛在这里会有所帮助。请不要将此问题标记为过于笼统，只需查看查询，您就会发现问题所在。

$term = mysqli_real_escape_string ($con , $_GET['term'] );
$location = mysqli_real_escape_string ($con , $_GET['location'] );
$keyword =  mysqli_real_escape_string ($con , $_GET['keyword'] );

if(empty($term)) {
$term1 = "";
}else{$term1 = "job_title LIKE '%".$term."%'";}

if(empty($location)) {
$loc1 = "";
}else{$loc1 = "location LIKE '%".$location."%'";}

if(empty($keyword)) {
$key1 = "";
}else{$key1 = "OR description LIKE '%".$keyword."%'";}


$sql = "SELECT * FROM jobs WHERE ".$term1." ".$loc1." ".$key1." ";
$r_query = mysqli_query($con,$sql);


while ($joblist = mysqli_fetch_array($r_query)){
$now = date('Y-m-d',time());

Answer 1

您在第一个 like 语句之后缺少 OR，这里 -

if(empty($location)) {
$loc1 = "";
}else{$loc1 = "location LIKE '%".$location."%'";}

您可能想在location like 之前放置一个OR，所以-

"OR location LIKE '%".$location."%'"

Answer 2

这是修改后的方法。试一试，如果它有效或无效，请回帖。

if (!empty($_GET['term'])) {
    $where[] = " job_title like ? ";
    $params[] = '%' . $_GET['term'] . '%';
}
if (!empty($_GET['location'])) {
    $where[] = " location like ? ";
    $params[] = '%' . $_GET['location'] . '%';
}
if (!empty($_GET['keyword'])) {
    $where[] = " description like ? ";
    $params[] = '%' . $_GET['keyword'] . '%';
}
$sql_where = !empty($where) ? ' where ' . implode(' or ', $where) : '';
$query = "SELECT * FROM jobs $sql_where";
if (!($tot = mysqli_prepare($con, $query))) {
   echo "Prepare failed: (" . mysqli_errno($con) . ") " . mysqli_error($con);
} else {
    if(!empty($params)) {
    $params = array_merge(array($tot),
        array(str_repeat('s', count($params))), 
        $params);
    call_user_func_array('mysqli_stmt_bind_param', $params);
    // adapated from http://stackoverflow.com/questions/793471/use-one-bind-param-with-variable-number-of-input-vars and http://www.pontikis.net/blog/dynamically-bind_param-array-mysqli may need to be altered
    }
    mysqli_execute($tot);
}

Answer 3

这似乎根据@chris85 的建议起作用。我不太确定 $params[] 是如何发挥作用的。任何解释都会很好。

if (!empty($_GET['term'])) {
$where[] = " job_title LIKE '%".$_GET['term']."%' ";
$params[] = '%' . $_GET['term'] . '%';
}
if (!empty($_GET['location'])) {
$where[] = " location LIKE '%".$_GET['location']."%' ";
$params[] = '%' . $_GET['location'] . '%';
}
if (!empty($_GET['keyword'])) {
$where[] = " description LIKE '%".$_GET['keyword']."%' ";
$params[] = '%' . $_GET['keyword'] . '%';
}
$sql_where = !empty($where) ? ' where ' . implode(' or ', $where) : '';
$query = mysqli_query ($con, "SELECT * FROM jobs $sql_where");