使用php搜索無法正常工作

Question

我正在根據產品名稱進行搜索。它確實有效，它確實根據搜索找到了該產品，但問題是，當我有幾個同名的產品時，它只顯示第一個產品。它如何優化？我很感謝你的幫助。

<?php 

//DATA coming from the browser 
$sSearch = $_GET['search']; 
//TURN it into UPPERCASE 
strtoupper($sSearch); 

//GETTING FROM FILE: 
$sajProducts = file_get_contents('products.txt'); 
$ajProducts = json_decode($sajProducts); 

$matchfound = false; 
//LOOPING THROUGH THE ARRAY OF PRODUCTS 
for ($i=0; $i< count($ajProducts); $i++) { 

    if ($sSearch == $ajProducts[$i]->name) { 
     $jSearchResult = $ajProducts[$i]; 
     $matchfound = true;  
     break; 
    } 
} 
//if there is a match display the product 
if ($matchfound) { 
    echo json_encode ($jSearchResult); 
    exit; 
} 
//if not display ALL products 
else { 

    echo json_encode($ajProducts); 
    exit; 
} 

?> 

而且JSON products.text文件：

[ 
    { 
     "id": "59d278cae7017", 
     "name": "A", 
     "price": "1", 
     "quantity": 3, 
     "image": "img_webshop\/productimage-59d74304917c2.jpg" 
    }, 
    { 
     "id": "59d27e20c8028", 
     "name": "A", 
     "price": "2", 
     "quantity": 3, 
     "image": "img_webshop\/productimage-59d743233c0cf.jpg" 
    }, 
    { 
     "id": "59d6a7ae16d15", 
     "name": "A", 
     "price": "3", 
     "quantity": 2, 
     "image": "img_webshop\/productimage-59d743392fbb5.jpg" 
    }, 
    { 
     "id": "59d6d6ee5f752", 
     "name": "A", 
     "price": "4", 
     "quantity": 4, 
     "image": "img_webshop\/productimage-59d74352d5b94.jpg" 
    }, 
    { 
     "id": "59d743d207bd5", 
     "name": "B", 
     "price": "5", 
     "quantity": 1, 
     "image": "img_webshop\/productimage-59d743d1e6e64.jpg" 
    }, 
    { 
     "id": "59d74451225ac", 
     "name": "B", 
     "price": "6", 
     "quantity": 1, 
     "image": "img_webshop\/productimage-59d7445120871.jpg" 
    }, 
    { 
     "id": "59e0d95cd4111", 
     "name": "B", 
     "price": "4", 
     "quantity": "5", 
     "image": "img_webshop\/productimage-59e0d95cd2c4b.jpg" 
    }, 
    { 
     "id": "59e0d992d1f3b", 
     "name": "C", 
     "price": "6", 
     "quantity": "5", 
     "image": "img_webshop\/productimage-59e0d992d19be.jpg" 
    }, 
    { 
     "id": "59e0d9c59fbf2", 
     "name": "D", 
     "price": "4", 
     "quantity": "5", 
     "image": "img_webshop\/productimage-59e0d9c59f1a5.jpg" 
    } 
] 

Answer 1

收集數組中的所有匹配結果，然後用JSON返回。

$jSearchResult = array(); 
for ($i=0; $i< count($ajProducts); $i++) { 

    if ($sSearch == $ajProducts[$i]->name) { 
     $jSearchResult[] = $ajProducts[$i]; 
     $matchfound = true;  
     //break; 
    } 
} 

因此，foreach循環完成了所有的JSON值迭代後，在這裏，我們將所有的比賽的最後陣列中$jSearchResult

然後返回它：

if ($matchfound) { 
    echo json_encode ($jSearchResult); 
    exit; 
} 

Answer 2

使用此代碼：

$matchfound = false; 
 
$jSearchResult = array(); 
 
//LOOPING THROUGH THE ARRAY OF PRODUCTS 
 
for ($i=0; $i< count($ajProducts); $i++) { 
 

 
    if ($sSearch == $ajProducts[$i]->name) { 
 
     $jSearchResult[] = $ajProducts[$i]; 
 
     $matchfound = true; 
 
    } 
 
}

取而代之的是：

$matchfound = false; 
 
//LOOPING THROUGH THE ARRAY OF PRODUCTS 
 
for ($i=0; $i< count($ajProducts); $i++) { 
 

 
    if ($sSearch == $ajProducts[$i]->name) { 
 
     $jSearchResult = $ajProducts[$i]; 
 
     $matchfound = true;  
 
     break; 
 
    } 
 
}