Php搜索腳本只拾取第一個字

作爲問題，它只拾取了我的product_name中第一個以var_char（50）作爲主鍵的單詞。Php搜索腳本只拾取第一個字

<?php 
error_reporting(E_ALL); 
ini_set('display_errors', '1'); 
$search_output = ""; 
if(isset($_POST['searchquery']) && $_POST['searchquery'] != ""){ 
    $searchquery = $_POST['searchquery']; 
    if($_POST['filter1'] == "Whole Site"){ 
    $sqlCommand = "SELECT * FROM products WHERE MATCH (product_name ,details,category,subcategory) AGAINST ('".$searchquery."' IN BOOLEAN MODE)"; 
    } 
    require_once("storescripts/connect_to_mysqli.php"); 
    $query = mysqli_query($myConnection,$sqlCommand) or die(mysqli_error($myConnection)); 
    $count = mysqli_num_rows($query); 
    if($count > 1){ 
     $search_output .= "<hr />$count results for <strong>$searchquery</strong><hr />$sqlCommand<hr />"; 
     while($row = mysqli_fetch_array($query)){ 
       $id=$row["id"]; 
      $product_name = $row["product_name"]; 
        $details= $row["details"]; 
       $category=$row["category"]; 
       $subcategory=$row["subcategory"]; 
      $search_output .= "ID: $id <br/> Name: $product_name -<br/>$details<br />$category<br/>$subcategory<br/> 
<a href='product.php?id=$id'>link</a><br/> 

"; 
     } // close while 
    } else { 
     $search_output = "<hr />0 results for <strong>$searchquery</strong><hr />$sqlCommand"; 
    } 
} 
?> 
<html> 
<head> 
</head> 
<body> 
<h2>Search the Exercise Tables</h2> 
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
Search For: 
    <input name="searchquery" type="text" size="44" maxlength="88"> 
Within: 
<select name="filter1"> 
<option value="Whole Site">Whole Site</option> 

</select> 
<input name="myBtn" type="submit"> 
<br /> 
</form> 
<div> 
<?php echo $search_output; ?> 
</div> 
</body>

這是由於它是一個主鍵？它實際上不工作的細節，以及，它其實很奇怪，不工作的大部分話，但工作some..i我完全confused..it不作任何感官

<?php 
error_reporting(E_ALL); 
ini_set('display_errors', '1'); 
require_once("storescripts/connect_to_mysqli.php"); 
$search_output = ""; 

if(isset($_POST['searchquery']) && $_POST['searchquery'] != ""){ 
    $searchquery = $_POST['searchquery']; 
    if($_POST['filter1'] == "Whole Site"){ 
$sqlCommand = sprintf("SELECT * FROM products WHERE product_name LIKE '%s%%'", 
       mysqli_real_escape_string($myConnection, $searchquery)); 
    } 

    $query = mysqli_query($myConnection,$sqlCommand) or die(mysqli_error($myConnection)); 
    $count = mysqli_num_rows($query); 
    if($count > 1){ 
     $search_output .= "<hr />$count results for <strong>$searchquery</strong><hr />$sqlCommand<hr />"; 
     while($row = mysqli_fetch_array($query)){ 
       $id=$row["id"]; 
      $product_name = $row["product_name"]; 
        $details= $row["details"]; 
       $category=$row["category"]; 
       $subcategory=$row["subcategory"]; 
      $search_output .= "ID: $id <br/> Name: $product_name -<br/>$details<br />$category<br/>$subcategory<br/> 
<a href='product.php?id=$id'>link</a><br/> 

"; 
     } // close while 
    } else { 
     $search_output = "<hr />0 results for <strong>$searchquery</strong><hr />$sqlCommand"; 
    } 
} 





?> 
<html> 
<head> 
</head> 
<body> 
<h2>Search the Exercise Tables</h2> 
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> 
Search For: 
    <input name="searchquery" type="text" size="44" maxlength="88"> 
Within: 
<select name="filter1"> 
<option value="Whole Site">Whole Site</option> 

</select> 
<input name="myBtn" type="submit"> 
<br /> 
</form> 
<div> 
<?php echo $search_output; ?> 
</div> 
</body> 
</html>

我上面的代碼試圖以及並縮小到product_name，它仍然不顯示，例如我有這個product_name作爲'歐米加魚'輸入，如果我輸入'魚'它不顯示一件事，如果我鍵入'梅加'那麼它的作品

我已經給了這個和LIKE功能試用，它似乎更好，但仍然有趣..請參考此鏈接對於那些有興趣的人

PHP search script for mySQL database, only 3 letter working

感謝

來源

2013-08-12 Philip Tiong

你調試的是？你確定查詢本身提供了錯誤的結果（並且沒有問題，例如，在用戶輸入中）？ –

如果$ _POST ['filter1']！=「Whole Site」，那麼$ sqlCommand爲空，因爲您沒有設置替代$ sqlCommand語句。 – jeff

你檢查了結果頁面的html源代碼嗎？ – Cups