Question

我们正在尝试使用7个搜索条件为具有8个属性的数据库执行搜索表单。但我们只希望一次搜索一个事件。这是我到目前为止的代码，并希望将搜索到的信息显示在表中。任何知道在哪里寻找的帮助都会受到赞赏。

<?php
include 'database_connector.php';
if(isset($_POST['submit'])){
 $type = $_POST['type'];
 $team1 = $_POST['team1'];
 $team2 = $_POST['team2'];
 $place = $_POST['place'];
 $year = $_POST['year'];
 $month = $_POST['month'];
 $day = $_POST['day'];
 $price = $_POST['price'];
 $date = $year.'-'.$month.'-'.$day;
  if($type)(
    $result=mysqli_connect($con, "select * from Sports where `Event Type` = '$type'")
    );
if($team1)(
$result1=mysqli_connect($con, "select * from Sports where `Team 1` = '$team1'")
);
if($team2)(
$result2=mysqli_connect($con, "select * from Sports where `Team 2` = '$team2'")
);  
if($place)(
$result3=mysqli_connect($con, "select * from Sports where `Place` = '$place'")
);
if($date)(
$result4=mysqli_connect($con, "select * from Sports where `Date` = '$date'")
);
if($price)(
$result5=mysqli_connect($con, "select * from Sports where `Price` = '$price'")
);
}
?>

Answer 1

使用if/elseif/只执行一个查询，并将结果分配给同一个变量：

if ($type) {
    $query = "select * from Sports where `Event Type` = '$type'";
} elseif ($team1) {
    $query = "select * from Sports where `Team 1` = '$team1'";
} ...
} else {
    die("You must fill in one of the search fields");
}

$result = mysqli_query($con, $query);

while ($row = mysqli_fetch_assoc($result)) {
    // Code to display each row of results
}

Answer 2

如果只想执行一个语句if/elseif/

，请使用相同的变量

if($type){
    $result=mysqli_connect($con, "select * from Sports where `Event Type` = '$type'")
    }
elseif($team1){
$result=mysqli_connect($con, "select * from Sports where `Team 1` = '$team1'")
}
elseif($team2){
$result=mysqli_connect($con, "select * from Sports where `Team 2` = '$team2'")
}

搜索结果不显示PHP MySql

2 个答案: