如何在php中找到2个不同的字符串（string1或string2）

Question

ValueError: No gradients provided for any variable: ['sequential/dense/kernel:0', 'sequential/dense/bias:0', 'sequential/dense_1/kernel:0', 'sequential/dense_1/bias:0', 'sequential/dense_2/kernel:0', 'sequential/dense_2/bias:0', 'sequential/dense_3/kernel:0', 'sequential/dense_3/bias:0'].

我在这里填写了表单，现在我尝试在输入表单时查找输入是否包含[key1] <form action="" method="post"> <?php if($input !== null){ //if input is submitted then show message if(strpos($input, '[key1]') === true or strpos($input, 'key2') === true) { //find the input [key1] or [key2] echo " input contain [key1] or [key2] "; }else{ // input doesnt contain [key1] or [key2] echo " input didnt contain [key1] or [key2]"; } }else{ // showing form input echo " <input type='text' name='inputText' maxlength='100' data-length=100 class='form-control char-textarea' rows='3' placeholder='use [key1] or [key2]' required aria-invalid='false'/> <input type='file' name='fileToUpload' id='fileToUpload' ><br> <button type='submit' name='SubmitButton' class='btn btn-primary mr-1 mb-1'>Send</button> "; }?> </form> [key2]

or

但是结果是if(strpos($input, '[key1]') === true or strpos($input, 'key2') === true) 而不是echo " input didnt contain [key1] or [key2]";

Answer 1

strpos返回一个整数（如果找到字符串，则返回位置），或者返回false（如果找不到该字符串）。所以你的测试

if(strpos($input, '[key1]') === true or strpos($input, 'key2') === true)

将永远不会过去，因为整数或false永远不会是=== true。相反，更改测试以检查strpos的结果是否不是false：

if(strpos($input, '[key1]') !== false or strpos($input, 'key2') !== false)