Question

我正在尝试从数据库构建select标签但是当我返回字符串时我得到了NULL。请帮助：

PHP代码

<?php       
function getDropDown($table,$value,$display){

    $dbConx = new mysqli("localhost", "root", "", "fw_request");

    if ($dbConx->connect_errno) {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }

    /* Select queries return a resultset */

    $sql = "SELECT * FROM `$table`";

    $result = $dbConx->query($sql);

    $a="<select name='streams'>";   
    $a="<option value=\"0\">Choose Stream</option>";


    while($row = $result->fetch_row()) {
        $a="<option value=\"1\" selected>1</option>";   
        $a='<br>';  
    }
    $a="</select>"; 
    return $a;
}

HTML代码

Sterams:<?php echo getDropDown('streams','Streams_ID','Streams_ShortName');

查询没问题。

谢谢， CFIR。

Answer 1

你继续覆盖$ a，你需要做的是通过连接 -

添加到$ a

$a = "<select name='streams'>";   
$a .= "<option value=\"0\">Choose Stream</option>";
while($row = $result->fetch_row()) {
   $a .= "<option value=\"1\" selected>1</option>";   
}
$a .= "</select>"; 
return $a;

Answer 2

您没有连接$ a的值。

这样做

function getDropDown($table, $value, $display) {

    $dbConx = new mysqli("localhost", "root", "", "fw_request");

    if ($dbConx->connect_errno) {
        echo "Failed to connect to MySQL: (" . $dbConx->connect_errno . ") " . $dbConx->connect_error;
    }

    /* Select queries return a resultset */

    $sql = "SELECT * FROM `$table`";

    $result = $dbConx->query($sql);

    $a = "<select name='streams'>";
    $a .= "<option value=\"0\">Choose Stream</option>";


    while ($row = $result->fetch_row()) {


        $a .= "<option value=\"1\" selected>1</option>";


    }
    $a .= "</select>";
    return $a;
}

从数据库构建<select> </select>

2 个答案: