如果其中一个类型值以字母p开头,我想显示是,但是我一直输出NONO。任何帮助都非常感谢...`
<?php
$pals[] = array('name' => 'jon' , 'type' => 'peach');
$pals[] = array('name' => 'gore' , 'type' => 'choc');
foreach($pals as $key => $value) {
if (substr($value['type'], 0) === "p") {
echo "Yes";
} else {
echo "NO";
}
}
答案 0 :(得分:2)
更新此行:
if (substr($value['type'], 0,1) === "p") {
答案 1 :(得分:1)
尝试substr($ value [&#39; type&#39;],0,1)
答案 2 :(得分:0)
如果你想要
,可以没有substr<?php
$pals[] = array('name' => 'jon' , 'type' => 'peach');
$pals[] = array('name' => 'gore' , 'type' => 'choc');
foreach($pals as $key => $value) {
$type = trim($value["type"]);
if ($type[0] === "p") {
echo "$type : Yes\n";
} else {
echo "$type : NO";
}
}
?>