Php/MySQL if语句，根据所选值返回信息

[page1.php]

<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>

我试图做的是在page1.php上，用户搜索关键字，然后从下拉列表中选择a、B或C。根据A、B、C，用户转到result.php，然后根据查询给出信息

---

上面的代码似乎不起作用[我根本没有得到任何结果][空白]。请提供帮助。

如果要搜索以字符串结尾的内容，必须将%通配符放在开头：

} 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>";
    }
}

---

您错过了LIKE子句之后的%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>";
}
}

---

请解释“似乎不起作用”-您收到错误消息了吗，您收到了什么？@user574632请参阅editAddErrorCheckingtoyourMysqli_查询调用。您的查询显示EndsWith实际上在搜索BeginsWith。@Barmar或diemysqli_error；你是这个意思吗？