php - 使用 php 到 mysql 的空字段

Question

我有在数据库中的一个字段中保存 3 个文本框的代码

当我输入 3 个文本框时没问题，但是当我填写 1 个文本框并按确定时将数据库中的另一个文本框保存为空白

我只想把文本框填满而忽略文本框为空

这是我的代码

<?php

include("connect.php");


$expert_name  = trim($_POST['expert_name']);
$expert_name2 = trim($_POST['expert_name2']);
$expert_name3 = trim($_POST['expert_name3']);


// this is for arabic language.
mysql_query("SET NAMES utf8");

// Insert data into mysql 
$sql="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
$sql2="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
$sql3="INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
$result=mysql_query($sql);
$result2=mysql_query($sql2);
$result3=mysql_query($sql3);
// if successfully insert data into database, displays message "Successful". 
if($result){
echo "Successful";
echo "<BR>";
// echo "<a href='formadd.php'>Back to main page</a>";
}
else {
echo "ERROR";
echo "<br>";
// this for print error in insert process 
echo mysql_error();
echo "<a href='expert_add.php'><br>Please try again </a>";
}
//mysql_close($con);
?>
<a href="<?php echo $_POST['sys']; ?>">back to form add</a>

Answer 1

执行您的sql 仅查询变量值不等于空。试试这个，

        $expert_name = trim($_POST['expert_name']);
        $expert_name2 = trim($_POST['expert_name2']);
        $expert_name3 = trim($_POST['expert_name3']);


        // this is for arabic language.
        mysql_query("SET NAMES utf8");

        // Insert data into mysql 
        if ($expert_name != "") {
            $sql = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name')";
            $result = mysql_query($sql);
        }

        if ($expert_name2 != "") {
            $sql2 = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name2')";
            $result2 = mysql_query($sql2);
        }
        if ($expert_name != "") {
            $sql3 = "INSERT INTO experts(id,expert_name) VALUES(NULL, '$expert_name3')";
            $result3 = mysql_query($sql3);
        }



        // if successfully insert data into database, displays message "Successful". 
        if ($result || $result2 || $result3) {
            echo "Successful";
            echo "<BR>";
            // echo "<a href='formadd.php'>Back to main page</a>";
        } else {
            echo "ERROR";
            echo "<br>";
            // this for print error in insert process 
            echo mysql_error();
            echo "<a href='expert_add.php'><br>Please try again </a>";
        }
        //mysql_close($con);
        ?>
        <a href="<?php echo $_POST['sys']; ?>">back to form add</a>

您还应该检查 $result2 和 $result3。我在这个答案中添加了这一点