Question

我想使用PHP和SQL查询几个表，并在单个HTML页面中返回格式。

这就是我现在所拥有的：

HTML：

<form name="searchForm" method="POST" action=""> <!-- My new PHP file would go here -->
  Search for: <input type="text" name="aSearch"> 
  <input type="submit" name="searchButton" value="Search">
</form>

控制器1（PHP）：

// Connect to DB
$servername = "";
$username = "";
$password = "";
$dbname = "";

// Create connection
$conn = new MySQLi($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}


// Collect Data phase:

// If an input has been given remove unnecessary characters
if(isset($_POST["aSearch"])) {
  $searchq = $_POST["aSearch"]; //searchq allows the input to search for part of a word

  $searchq = preg_replace("#[^0-9a-z]#i","",$searchq); //Can only search words

// Select statements if the searched word and insectName match
$sql = "SELECT * FROM insectt WHERE insectName LIKE '%$searchq%'";
}

// Tests if the code been inserted
if ($conn->query($sql)=== TRUE){
  echo "The rows you have searched for are:";
} else {
  echo "Connection failed: ";
  echo $conn->error;
}

// Show fields
$result = $conn->query($sql);

// Output data of each row
if ($result-> num_rows> 0) {
  readfile("ViewReturn.html"); //Output it in an html file
  while($row = $result-> fetch_assoc()) {
    echo "Insect Name: ".$row["insectName"]. "<br><br>";
  }
} else {
  echo "0 results";
}

// Close connections
if (!function_exists('mysqli_init') && !extension_loaded('mysqli')) {
  echo "There was an error with MySQLi";
} else {
  echo "Closing Connections";
}

$conn-> close();
exit();
?>

控制器2（PHP）：

// Connect to DB
$servername = "";
$username = "";
$password = "";
$dbname = "";

// Create connection
$conn = new MySQLi($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}

// Collect Data phase:

// If an input has been given remove unnecessary characters
if(isset($_POST["aSearch"])) {
  $searchq = $_POST["aSearch"]; //searchq allows the input to search for part of a word
  $searchq = preg_replace("#[^0-9a-z]#i","",$searchq); //Can only search words

// Select statements if searched
$sql = "SELECT * FROM birdt WHERE birdName LIKE '%$searchq%'";
}

// Tests if the code been inserted
if ($conn->query($sql)=== TRUE){
  echo "The rows you have searched for are:";
} else {
  echo "Connection failed: ";
  echo $conn->error;
}

// Show fields
$result = $conn->query($sql);

// Output data of each row
if ($result-> num_rows> 0) {
  readfile("ViewReturn.html"); //Output it in an html file
  while($row = $result-> fetch_assoc()) {
    echo "Bird Name: ".$row["birdName"]. "<br><br>";
  }
} else {
  echo "0 results";
}

// Close connections
if (!function_exists('mysqli_init') && !extension_loaded('mysqli')) {
  echo "There was an error with MySQLi";
} else {
  echo "Closing Connections";
}

$conn-> close();
exit();
?>

SQL表：

birdt:                    insectt:
╔════════╦═══════════╗    ╔══════════╦════════════╗ 
║ birdID ║  birdName ║    ║ insectID ║ insectName ║
╠════════╬═══════════╣    ╠══════════╬════════════╣
║   1    ║   Gull    ║    ║    1     ║   cricket  ║
║   2    ║   Hawk    ║    ║    2     ║   bee      ║
║   3    ║   Pigeon  ║    ║    3     ║   flea     ║
╚════════╩═══════════╝    ╚══════════╩════════════╝

目前，我可以通过更改HTML表单中的操作（我可以从下拉菜单中执行）单独搜索两个表，并在不同的HTML页面中单独返回查询结果，但这就是它。

我想知道的是，是否有办法从单个控制器查询两个SQL表（并使我的html表单中的新控制器成为可能），这可以查询我的两个表并返回搜索结果在单个HTML页面中。因此，如果我要搜索与鸟类或插入表中的记录名称匹配的单个关键字 - 或两者 - 它将返回它。

可能需要更改的是我的PHP文件中的SQL查询，但我不确定要将其更改为。

我真的不知道如何解决这个问题，网上的信息很混乱，所以我会接受任何帮助。

提前感谢您的帮助。：¬）

Answer 1

您可以使用多个查询

$sql = "SELECT * FROM birdt WHERE birdName LIKE '%$searchq%';";
$sql .= "SELECT * FROM insectt WHERE insectName LIKE '%$searchq%'";

并使用mysql_multi_query()

执行

更多访问：http://www.w3schools.com/php/func_mysqli_multi_query.asp

查询多个SQL表并从单个HTML页面中的一个或多个表返回结果

1 个答案: