Question

我有以下代码，其中内部图像标签我打开php标签，并根据ifelse（）条件决定不同的图像。我想为所选图像添加不同的标题但无法提供解决方案。代码如下：

<img title = "Last assessment for this child was submitted <?php if ($time == 0){echo $time;}else{echo $time - 1;}?> Month(s) ago." 
            src="<?php 
                            if ($record->$period == 0) { echo base_url()."img/warning.png";}
                            else{

                                date("M d, Y", strtotime($record->$period));
                                $vtime = new DateTime($record->$period);             ///////////////////////
                                $today = new DateTime(); // for testing purposes            ///Calculate  Time period//
                                $diff = $today->diff($vtime);                                   ///
                                $time = $diff -> m;
                                if($time <= 4)
                                {echo base_url()."img/green.png";}
                                elseif( $time > 4 && $time <= 6)
                                {echo base_url()."img/yellow.png";}
                                elseif($time >= 6)
                                {echo base_url()."img/red.png";}
                            }
                " 
    />

我想要第一个条件的不同标题。即如果第一个条件为真且所示图像为“warning.png”。然后图像标题应该是“检查记录”而不是标题“提交的最后一次评估是......”

非常感谢任何帮助。

Answer 1

您可以简单地使用以下内容。只需将当前if ... else条件嵌套在title标签上的另一个if ... else条件中。

<img title = "
    <?php if($record->period==0) 
        echo "Check record"; 
    else { ?>
        Last assessment for this child was submitted <?php if ($time == 0){echo $time;}else{echo $time - 1;}?> Month(s) ago.
    <?php } ?>" 
            src="<?php 
                            if ($record->$period == 0) { echo base_url()."img/warning.png";}
                            else{

                                date("M d, Y", strtotime($record->$period));
                                $vtime = new DateTime($record->$period);             ///////////////////////
                                $today = new DateTime(); // for testing purposes            ///Calculate  Time period//
                                $diff = $today->diff($vtime);                                   ///
                                $time = $diff -> m;
                                if($time <= 4)
                                {echo base_url()."img/green.png";}
                                elseif( $time > 4 && $time <= 6)
                                {echo base_url()."img/yellow.png";}
                                elseif($time >= 6)
                                {echo base_url()."img/red.png";}
                            }
                " 
    />

Answer 2

就像我在评论中所说的那样，你可以将逻辑分开，在这里和那里进行计算。完成后，设置变量，然后在演示文稿中将其回显：

<?php
// initialization
$title = '';
$src = '';

// logic
$time = ($time == 0) ? $time : $time - 1;
$title = "Last assessment for this child was submitted %s Month(s) ago."; // initial

if ($record->$period == 0) { 
    $src = base_url() . "img/warning.png";
    // override $title
    $title = 'Check record';
} else {
    $vtime = new DateTime($record->$period);            
    $today = new DateTime();
    $diff = $today->diff($vtime);
    $time = $diff->m;
    if($time <= 4) {echo ;
        $src = base_url()."img/green.png";
    } elseif( $time > 4 && $time <= 6) {
        $src = base_url()."img/yellow.png";
    } elseif($time >= 6) {
        $src = base_url()."img/red.png";
    } else {
        // whatever you need to do
    }
}

$title = sprintf($title, $time);

?>
<!-- HTML MARKUP -->
<img title="<?php echo $title; ?>" src="<?php echo $src; ?>" />

Answer 3

你应该避免在HTML中插入如此多的PHP代码以保持代码的可读性。

if ($record->period === 0) {
    echo '<img src="img/warning.png" title="Warning title" />';
} else {
    // Are you sure this does what you want?
    // You probably need $record->period. (no $)
    $vtime = new DateTime($record->$period);
    $today = new DateTime();
    $diff = $today->diff($vtime);
    $time = $diff->m;
    $title = 'Last assessment for this child was submitted ' . 
        ($time === 0 ? 0 : $time-1) . 
        ' month(s) ago.';

    if ($time <= 4) {
        echo '<img src="img/green.png" title="'. $title . '" />';
    } else if ($time <= 6) {    
    // Don't need to check if it's bigger than 4, you've already checked this 
    // in the initial "if" and if that was succesful, we wouldn't be here.
        echo '<img src="img/yellow.png" title="'. $title . '" />';
    } else {
        echo '<img src="img/red.png" title="' . $title . '" />';
    }
}

Answer 4

PHP很容易嵌入HTML。所以使用它。

$imageURL = "";
if ($record->$period == 0) {
    $imageURL = base_url() . "img/warning.png";
} else {
    // date("M d, Y", strtotime($record->$period)); // This is not usable remove this statement
    $vtime = new DateTime($record->$period);             ///////////////////////
    $today = new DateTime(); // for testing purposes            ///Calculate  Time period//
    $diff = $today->diff($vtime);                                   ///
    $month = $diff->m;
    if ($month <= 4) {
        $imageURL = base_url() . "img/green.png";
    } elseif ($month > 4 && $month <= 6) {
        $imageURL = base_url() . "img/yellow.png";
    } else {
        $imageURL = base_url() . "img/red.png";
    }
}

重写您的图片代码，如下所示：

<img title = "Last assessment for this child was submitted <?php echo ($time)? $time - 1: $time; ?> Month(s) ago." 
     src="<?php echo $imageURL; ?>" />

您的代码存在一些错误。请仔细检查以下几点以获取进一步的发展。

date("M d, Y", strtotime($record->$period))：为什么这个陈述存在，因为你没有保存日期函数保存的结果。

$vtime :什么是vtime？变量应该是简短而有意义的名称。

$diff->m :这将返回月份编号，以便更具体，而不是$time使用$month作为变量名来存储月份值。

关于if条件

首先If：检查$month is <= 4：正确
秒If：检查$month > 4 or <=6
第三个elseif：在旧代码中。为什么我们需要这个，因为如果不是满足以上两个条件则意味着它是强制性的>= 6 然后将其直接放入其他部分。

使用图像的替代标题

4 个答案: