php-如果存在则显示图像字符串,如果字段为空则显示默认图像

像往常一样,我在使用PHP时遇到了麻烦:(我对此仍然很陌生,我希望最终它会变得更容易…

这真让我头疼,希望有人能帮助您.

我正在尝试从数据库列中提取一串图像.如果存在图像,则将显示它们,否则默认的“无图像”图像将以相同的格式显示.

甚至不确定我是否会完全正确.数据库列/行中可以有多个值吗？这是一个数组吗？

到目前为止,这是我的代码…

$images_array = "SELECT images FROM properties WHERE property = '".$property."' "; // set images query 
$images = mysql_query($images_array); // run images_array query save as images
$images_count = mysql_num_rows($images);
if ($images_count > 0) {
while ($image = mysql_fetch_array($images)) {
echo'<div id="property_images">
<div id="property_main_image"><img alt="Property Image" src="./images/properties/'.$image['images'].'" /></div>
<div id="property_sub_image"><img alt="Property Image" src="./images/properties/'.$image['images'].'" /></div>
<div id="property_sub_image"><img alt="Property Image" src="./images/properties/'.$image['images'].'" /></div>
</div>';
}
} else {
echo'<div id="property_images">
<div id="property_main_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
<div id="property_sub_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
<div id="property_sub_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
</div>';
}

当数据库中存在图像时,它将正确显示图像,否则将显示alt =“属性图像”.这使我认为代码始终将$images视为TRUE …？

我可以通过保存多个具有相同MySQL值的文件名并获取数组/字符串来以此方式显示多个图像吗？

在此先感谢您的帮助:)

根据要求我的数据库结构…

| properties
| ID | address | area | postcode | phone | mobile | tenancy | type | available | deposit | rent | description | bedrooms | bathrooms | communal | kitchens | parking | garden | broadband | property | vacancy | images |

解决方法:

执行查询时,$image变量始终返回TRUE.

因此,我的建议是执行以下操作：

$count = mysql_num_rows($query);
if ($count > 0 ) {
....
}

更新：

$images = mysql_query($images_array); // run images_array query save as images
$images_count = mysql_num_rows($images);

while ($image = mysql_fetch_array($images)) {
    $lettercount = strlen(str_replace(' ', '', $images));
    if($lettercount > 0) {
        /* show the images */
    }else{
        /* show the default images */
    }
}

更新2：

$images_array = "SELECT * FROM properties WHERE property = '".$property."' "; // set images query 
$images = mysql_query($images_array); // run images_array query save as images

while ($row = mysql_fetch_assoc($images)) { 
        $check_string_lenght = strlen(str_replace(' ', '', $row['images']));
    if($check_string_lenght > 0) {
    /* or if($row['images'] != '') { */
       echo'<div id="property_images">
       <div id="property_main_image"><img alt="Property Image" src="./images/properties/'.$row['images'].'" /></div>
       <div id="property_sub_image"><img alt="Property Image" src="./images/properties/'.$row['images'].'" /></div>
       <div id="property_sub_image"><img alt="Property Image" src="./images/properties/'.$row['images'].'" /></div>
       </div>';
} else {
       echo'<div id="property_images">
       <div id="property_main_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
       <div id="property_sub_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
       <div id="property_sub_image"><img alt="No Property Image" src="./images/properties/no_image" /></div>
       </div>';
    }

}

标签：if-statement,select,mysql,php
来源： https://codeday.me/bug/20191123/2065974.html