PHP/MySQL的，如果返回基於價值信息的聲明選擇

[page1.php中]PHP/MySQL的，如果返回基於價值信息的聲明選擇

<form action="result.php" method="POST" style='color:white;font-family: __custom;'> 
    <input type="text" placeholder="Search.." name="search" /><br><br> 
    <center> 
     <select name="example"> 
     <option value="A">A</option> 
     <option value="B">B</option> 
     <option value="C">C</option> 
     </select> 
     <br><br> 
    </center> 
    &nbsp;<input type="submit" value="GO!" id="sub"/> 
</form><br><br><br>

[result.php]

<?php 
$searchq = $_POST['search'] // search textbox; 
//Connect to mySQL and retrieve all contacts 
$conErr = "Error connecting to server or database"; 
$con = mysqli_connect("host", "user", "password", "database"); 
// Check connection 
if (mysqli_connect_errno()) { 
    echo $connErr; 
} 
if ($_POST['example'] == 'C') { 
    echo "You chose inCludes"; 
    $result = mysqli_query($con, "SELECT * FROM people WHERE name LIKE '%$searchq%' ORDER BY name ASC"); 
    while ($row = mysqli_fetch_array($result)) { 
     echo "<hr>"; 
     echo "All information regarding Select Statement"; 
     echo "<br>"; 
    } 
} else if ($_POST['example'] == 'A') { 
    echo "You chose EndsWith"; 
    $result = mysqli_query($con, "SELECT * FROM people WHERE name LIKE '$searchq%' ORDER BY name ASC"); 
    while ($row = mysqli_fetch_array($result)) { 
     echo "<hr>"; 
     echo "All information based on Select Statement"; 
     echo "<br>"; 
    } 
} else { 
    echo "Blah"; 
} 
mysqli_close($con); 
?> 
</div>

我想要做的是第1頁。 php用戶搜索關鍵字，從下拉列表中選擇A，B或C。基於A，B，C，用戶轉到result.php，然後根據查詢給出信息。

上面的代碼似乎沒有工作[我根本沒有得到任何結果] [空白]。請幫忙。

來源

2014-09-10 user4013633

請解釋'似乎沒有工作' - 你有錯誤信息，你會得到什麼？ – Steve 2014-09-10 21:40:36

@ user574632請參閱編輯 – user4013633 2014-09-10 21:41:41

將錯誤檢查添加到您的'mysqli_query（）'調用中。 – Barmar 2014-09-10 21:42:27

如果你要搜索的東西，用字符串結尾，你必須把%通配符開頭：

} else if ($_POST['example'] == 'A') { 
    echo "You chose EndsWith"; 
    $result = mysqli_query($con, "SELECT * FROM people WHERE name LIKE '%$searchq' ORDER BY name ASC"); 
    while ($row = mysqli_fetch_array($result)) { 
     echo "<hr>"; 
     echo "All information based on Select Statement"; 
     echo "<br>"; 
    } 
}

來源

2014-09-10 21:48:15 Barmar

完美謝謝！ – user4013633 2014-09-10 21:53:23

你喜歡的條款後無緣％通配符

else if ($_POST['example'] == 'A') { 
echo "You chose EndsWith"; 
$result = mysqli_query($con, "SELECT * FROM people WHERE name LIKE '%$searchq%' ORDER BY name ASC"); 
while ($row = mysqli_fetch_array($result)) { 
    echo "<hr>"; 
    echo "All information based on Select Statement"; 
    echo "<br>"; 
} 
}

來源

2014-09-10 22:13:16 Chirag

PHP/MySQL的，如果返回基於價值信息的聲明選擇

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